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clc; pathname=get_absolute_file_path('13_11_soln.sce') filename=pathname+filesep()+'13_11_data.sci' exec(filename) // Solution: // upstream temperature in Rankine, T1=T1+460; //deg R // absolute downstream pressure, p2=p2+14.7; //psia // flow capacity constant, Cv=(Q/22.7)*sqrt(T1/(p2*del_p)); // Results: printf("\n Results: ") printf("\n The flow capacity constant is %.2f.",Cv)
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clear; clc; //Atomic weigths Si=28.1 Ga=69.7 As=74.9 Na=6.02*10^23 // Avagadro Number in mol^-1 //(a)Si a=5.43*10^-8 //in cm n=8 //no. of atoms/cell //(b)GaAs a1=5.65*10^-8 //in cm //Calculation N=8/a^3 //Atomic Concentration in atoms/cc N1=4/a1^3 //Atomic Concentration in atoms/cc Density=(N*Si)/(Na) Density1=(N1*(Ga+As))/(Na) mprintf("Density of Si= %1.2f g/cm^3\n",Density) mprintf("Density of GaAs= %1.2f g/cm^3",Density1)
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//Scilab Code for Example 7.11 of Signals and systems by //P.Ramakrishna Rao clear; clc; clear x y n; x=[0,0,2,0,0]; y=[0,0,1,1,0]; n=-2:2; c = gca(); c.y_location = "origin"; c.x_location = "origin"; plot2d2(n,x,2); title('x(t)') xlabel('t') figure(1); n=-2:2; c = gca(); c.y_location = "origin"; c.x_location = "origin"; plot2d2(n,y,5); title('y(t)') xlabel('t') z=conv(x,y); figure(2); n=-3:5; c = gca(); c.y_location = "origin"; c.x_location = "origin"; plot(n,z,2); title('Convoluted signal z(t)') xlabel('t')
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//Exa 4.5 clc;clear;close; format('v',6); I=100;//A///Current V=11;//kV Xs=4;//ohm/phase f=50;//Hz pf=0.8;//Power factor Lagging //Calculation theta=acosd(pf);//degree disp("Part(a)"); E=V*1000/sqrt(3)+I*(cosd(theta)-%i*sind(theta))*%i*Xs;//V del=atand(imag(E)/real(E));//degree E=abs(E);//V/phase disp(E,"Open circuit phase emf(V/phase)"); disp(del,"Angle delta(degree)"); disp("Part(b)"); del_dash=10+del;//degree P_by_V=E*sind(del_dash)/Xs;//per phase //P=V*I*cos_fi I_cos_fi=P_by_V; //V*1000/sqrt(3)+I*(cos_fi-%i*sin_fi)*%i*Xs=E I_sin_fi={sqrt(E^2-(4*I_cos_fi^2))-V*1000/sqrt(3)}/4; tan_fi=I_sin_fi/I_cos_fi; fi=atand(tan_fi);//degree I=I_cos_fi/cosd(fi);//A disp(I,"New load current(A)"); pf=cosd(fi);//lagging power factor disp(pf,"Its power factor(lagging)"); disp("Part(c)"); pf1=0.8;///original power factor Idash=I*pf/pf1;//Current disp(Idash,"New value of load current(A)"); //Answer is slightly differ because of accuracy in calculations.
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// Updated(19-7-07) // 12.4 A=[1 -0.8]; dA=1; B=[0.4 0.6]; dB=1; rho = 0.8; k = 1; N1 = 0; N2 = 3; Nu = 2; getf gpc_N.sci; [K,KH1,KH2,Tc,dTc,Sc,dSc,R1,dR1] = ... gpc_N(A,dA,B,dB,k,N1,N2,Nu,rho)
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//example 8.1 clc; funcprot(0); // Initialization of Variable h1=2758.0; h2=1794.8; h3=173.88; h4=h3+1.0084/1000*(8-0.008)*1000; neta=(h1-h2-h4+h3)/(h1-h4); disp(neta*100,"thermal efficiency in %"); bwr=(h4-h3)/(h1-h2); disp(bwr*100,"back work ratio in %"); mdot=100*1000*3600/(h1-h2-h4+h3); disp(mdot,"mass flow rate in kg/h"); Qindot=mdot*(h1-h4)/3600/1000; disp(Qindot,"energy inflow rate in MW"); Qoutdot=mdot*(h2-h3)/3600/1000; disp(Qoutdot,"energy outflow rate in MW"); disp(Qoutdot/Qindot*100,"ratio of energy outflow/inflow in %"); mcwdot=mdot*(h2-h3)/(146.68-62.99); disp(mcwdot,"mass flow rate in kg/h"); clear()
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//Problem 40.09: Two parallel wires, each of diameter 5 mm, are uniformly spaced in air at a distance of 50 mm between centres. Determine the capacitance of the line if the total length is 200 m. //initializing the variables: e0 = 8.85E-12; er = 1; D = 0.05; // in m d = 0.005; // in m l = 200; // in m //calculation: //capacitance C = %pi*e0*er/(log(D/(d/2))) //capacitance of a 200 m length C200 = C*l printf("\n\n Result \n\n") printf("\n capacitance of a 200 m length is %.2E F",C200)
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-- VectorCAST 18.sp2 (07/02/18) -- Test Case Script -- -- Environment : AIRPORT -- Unit(s) Under Test: airport -- -- Script Features TEST.SCRIPT_FEATURE:C_DIRECT_ARRAY_INDEXING TEST.SCRIPT_FEATURE:CPP_CLASS_OBJECT_REVISION TEST.SCRIPT_FEATURE:MULTIPLE_UUT_SUPPORT TEST.SCRIPT_FEATURE:MIXED_CASE_NAMES TEST.SCRIPT_FEATURE:STATIC_HEADER_FUNCS_IN_UUTS --
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clc clear printf("Example 12.7 | Page number 422 \n\n"); //Find //(a) Actual air //(b) Excess air //(c) Percentage theoritical air //(d) Mass fractions //(e) Dew point //Part(a) printf("Part(a)\n") mC = 0.65 //kg //mass of C per kg coal mA = 0.15 //kg //mass of Ash per kg coal CR = 0.05 //kg //mass of carbon in solid refuse per kg coal mR = 0.2 //kg //mass of refuse per kg coal m = mC- CR //kg //mass of carbon burnt per kg coal //By C balance x = (14 + 1)*(12/0.6) //kg //mass of burnt coal //By H2 balance b = x*(0.06/2) //By O2 Balance a = (14 + 0.5 + 3.5 + 4.5)-(x*0.1/32) actual_air = a*(32+3.76*28)/x //kg/kg coal printf("Actual air = %.3f kg/kg coal\n\n",actual_air) //Part(b) printf("Part(b)\n") Stoichiometric_air = (0.6*11.45+0.06*34.3)-(0.1/0.232) //kg excess_air = (actual_air - Stoichiometric_air)/Stoichiometric_air*100 printf("Excess air = %.1f%%\n\n",excess_air) //Part(c) printf("Part(c)\n"); printf("Percentage theoritical air = %.1f%%\n\n",100+excess_air) //Part(d) printf("Part(d)\n") m = 14*44 + 1*28 +3.5*32 +81.5*28 +9*18 //kg //mass of combustion product printf("Mass fraction of CO2 = %.2f%%\n",14*44/m*100) printf("Mass fraction of CO = %.2f%%\n",1*28/m*100) printf("Mass fraction of O2 = %.2f%%\n",3.5*32/m*100) printf("Mass fraction of N2 = %.2f%%\n",81.5*28/m*100) printf("Mass fraction of H2O = %.2f%%\n\n",9*18/m*100) //Part(e) printf("Part(e)\n") xH2O = 9/(14+1+3.5+81.5+9) //molfraction of H2O pH2O = xH2O*1e5 //Pa //partial pressure //From steam table tdp = 42.5 //°C printf("Dew point temperature = %.1f °C",tdp)
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exec("swigtest.start", -1); if endif_get() <> 1 then swigtesterror(); end if define_get() <> 1 then swigtesterror(); end if defined_get() <> 1 then swigtesterror(); end if 2 * one_get() <> two_get() then swigtesterror(); end exec("swigtest.quit", -1);
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//Chapter-11, Example 11.9, Page 494 //============================================================================= clc clear //INPUT DATA T=100;//Temperature of dry steam in degree C Do=0.025;//Outer diameter of the pipe in m Ts=84;//Surface temmperature of pipe in degree C Tf=(T+Ts)/2;//Film temperature in degree C p1=963.4;//Density of liquid in kg/m^3 u=(306*10^-6);//Dynamic viscosity in N.s/m^2 hfg=2257;//Enthalpy in kJ/kg pv=0.596;//Density of vapour in kg/m^3 k1=0.677;//Thermal conductivity in W/m.K //CALCULATIONS h=(0.725*((9.81*p1*(p1-pv)*k1^3*hfg*1000)/(u*(T-Ts)*Do))^0.25);//Heat transfer coefficient in W/m^2.K q=(h*3.14*Do*(T-Ts))/1000;//Heat transfer per unit length in kW/m m=(q/hfg)*3600;//Total mass flow of condensate per unit length in kg/h //OUTPUT mprintf('Rate of formation of condensate per unit length is %3.2f kg/h',m) //=================================END OF PROGRAM==============================
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function [x,y,typ]=standard_outputs(o) //get position of inputs ports and clock inputs port for a standard block // the output ports are located on the right (or left if tilded) vertical // side of the block, regularly located from bottom to top // the clock output ports are located on the bottom horizontal side // of the block, regularly located from left to right //! xf=60 yf=40 graphics=o(2) model=o(3) orig=graphics(1);sz=graphics(2);orient=graphics(3); out=size(model(3),1);clkout=size(model(5),1); if orient then xo=orig(1)+sz(1) dx=xf/7 else xo=orig(1) dx=-xf/7 end // output port location if out==0 then x=[];y=[],typ=[] else y=orig(2)+sz(2)-(sz(2)/(out+1))*(1:out) x=(xo+dx)*ones(y) typ=ones(x) end // clock output port location if clkout<>0 then x=[x,orig(1)+(sz(1)/(clkout+1))*(1:clkout)] y=[y,(orig(2)-yf/7)*ones(1,clkout)] typ=[typ,-ones(1,clkout)] end
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function [ECmInv,EeV] = HCl(var,J) //Written by Aditi and O.S.K.S. Sastri //var is vector consisting of De, b and Re values. //Typical input for var for HCl is [5,1,1.27455] //J is rotational quantum number. //J = 0 gives pure vibrational levels //J = 1 gives energy eigen values corresponding to //first excited rotational for v = 0, 1, 2,... a0 = 6; // infinite square well width in Angstroms N0 = 150; //Number of basis functions [V,E10] = potential(J,var); //Defines the potential function h = hmatrix(a0,N0,V,E10); //Determines the hmatrix evals = spec(h); //spec command used for obtaining the eigen values EeV = gsort(evals); //gsort command used for sorting data in descending order ECmInv = EeV/(1239.84193*10^(-7));//energies in wavenumbers endfunction function [V,E10] = potential(J,var,a0) // Model parameters De = var(1); //Molecular dissociation energy expressed in eV b = var(2); //parameter in Morse potential definition expressed in Angstroms^(-1) Re = var(3); //equilibrium bond length in Angstroms R= 0.6:0.001:a0; // discretising distance parameter // Defining Morse potential v = De*((exp(-2*b*(R-Re)))-(2*exp(-b*(R-Re)))); mu = 0.9796*931.49410*10^6; // reduced mass of molecule in eV hbarc = 1973.29; //value is in eV-Angstroms //Defining centrifugal potential for rotational term vcf = (J*(J+1)*hbarc^2./(2*mu*R.^2)); V = v+vcf; plot(R,V); // ground state energy of infinite square well potential in eV E10 = (%pi^2*hc^2)/(2*mu*(a0^2)); endfunction function [h] = hmatrix(a0,N0,V,E10) h = zeros(N0,N0); for m = 1:N0 h(m,m) = m^2*E10 + Vmm(m,a0,V); //Diagonal elements for n = m+1:N0 h(m,n) = Vnm(n,m,a0,V); //Non-Diagonal elements h(n,m) = h(m,n); end end endfunction function [I1] = Vmm(m,a0,V) R = 0.6:0.001:a0; c1 = 1-cos(2*m*%pi*R/a0); f1 = c1.*V/a0; I1 = intsplin(R,f1); // Integration using spline interpolation endfunction function [I2] = Vnm(n,m,a0,V) R = 0.6:0.001:a0; c1 = cos((n-m)*%pi*R/a0); c2 = cos((n+m)*%pi*R/a0); f2 = V.*(c1-c2)/a0; I2 = intsplin(R,f2); //Integration using spline interpolation endfunction
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P=10000 V=1000 W1=4*%pi*10^6 Wc=2*%pi*10^8 a=P/V^2 printf("\na=%.2f",a) //(b)= A=1000+2*225+2*150+2*75 peak_power=a*A^2 printf("\nA=%.0f V\npeak_power=%.0f W",A,peak_power)
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//Exa 2.3 clc; clear; close; //given data n=10^24;//electrons/m^3 e=1.6*10^-19;//constant v=1.5*10^-2;//in m/s A=1;//in cm^2 A=1*10^-4;//in m^2 I=e*n*v*A;//in Ampere disp(I,"Magnitude of current in Ampere : ");
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//Problem 16.12: A motor has an output of 4.8 kW, an efficiency of 80% and a power factor of 0.625 lagging when operated from a 240 V, 50 Hz supply. It is required to improve the power factor to 0.95 lagging by connecting a capacitor in parallel with the motor. Determine (a) the current taken by the motor, (b) the supply current after power factor correction, (c) the current taken by the capacitor, (d) the capacitance of the capacitor, and (e) the kvar rating of the capacitor. //initializing the variables: Pout = 4800; // in Watt eff = 0.8;// effficiency f = 50; // in ohm V = 240; // in Volts pf1 = 0.625// power factor pf2 = 0.95// power factor //calculation: Pin = Pout/eff Im = Pin/(V*pf1) phi1 = acos(pf1) phi1d = phi1*180/%pi //When a capacitor C is connected in parallel with the motor a current Ic flows which leads V by 90°. phi2 = acos(pf2) phi2d = phi2*180/%pi Imh = Im*cos(phi1) //Ih = I*cos(phi2) Ih = Imh I = Ih/cos(phi2) Imv = Im*sin(phi1) Iv = I*sin(phi2) Ic = Imv - Iv C = Ic/(2*%pi*f*V) kvar = V*Ic/1000 printf("\n\n Result \n\n") printf("\n (a)current taken by the motor, Im = %.0f A",Im) printf("\n (b)supply current after p.f. correction, I = %.2f A ",I) printf("\n (c)magnitude of the capacitor current Ic = %.0f A",Ic) printf("\n (d)capacitance, C = %.0f μF ",(C/1E-6)) printf("\n (d)kvar rating of the capacitor = %.2f kvar ",kvar)
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clc //initialisation of variables clear W= 38 //rev/sec w= 62.4 //lbf/ft^3 m= 2000 //lbm/sec g= 32.2 //ft/sec^2 ps= 5000 //lbf/ft^2 S3= 4.6 e= 0.91 //CALCULATIONS S1= W*(w*m^2/(g*ps)^3)^0.25 D= S3*(m^2/(w*g*ps))^0.25 //RESULTS printf ('S1 = %.3f',S1) printf ('\n Diameter = %.2f ft',D) printf ('\n efficiency = %.2f ',e)
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//no i/p args are passed to the function X=corrmtx(); //output //!--error 4 //Undefined variable: varargin //at line 88 of function corrmtx called by : //X=corrmtx();
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clc; clear; h=6.63*10^-34 //Plancks constant in J-s c=3*10^8 //velocity of light in m/s m=9.1*10^-31 //mass in kg lambda_1=100*10^-12 //wavelength in m e=1.6*10^-19 //charge in C //calculation delta_lambda=(h/(m*c)) //wavelength in m mprintf("The compton shift is = %1.2e m\n",delta_lambda) lambda_0=lambda_1-delta_lambda //wavelength of the scattered photon in m delta_E=(h*c*delta_lambda)/(lambda_1*lambda_0) mprintf("\nThe kinetic energy imparted to the electron is = %1.2e J or %1.2f eV",delta_E,delta_E/e) //The answer provided in the textbook is wrong.
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function z=zeros(n,m) // Copyright INRIA [lhs,rhs]=argn(0) if rhs==1 then z=0*ones(n);return;end if rhs==2 then z=0*ones(n,m);return;end
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//Problem 7.09: //initializing the variables: F1 = 50000; // in lb/h F2 = 60000; // in lb/h F3 = 80000; // in lb/h F4 = 60000; // in lb/h F5 = 40000; // in lb/h F6 = 35000; // in lb/h Cp1 = 0.65; // in Btu/lb.degF Cp2 = 0.58; // in Btu/lb.degF Cp3 = 0.78; // in Btu/lb.degF Cp4 = 0.70; // in Btu/lb.degF Cp5 = 0.52; // in Btu/lb.degF Cp6 = 0.60; // in Btu/lb.degF Tin1 = 70; // in deg F Tin2 = 120; // in deg F Tin3 = 90; // in deg F Tin4 = 420; // in deg F Tin5 = 300; // in deg F Tin6 = 240; // in deg F Tout1 = 300; // in deg F Tout2 = 310; // in deg F Tout3 = 250; // in deg F Tout4 = 120; // in deg F Tout5 = 100; // in deg F Tout6 = 90; // in deg F //calculation: Duty1 = F1*Cp1*(Tout1 - Tin1) Duty2 = F2*Cp2*(Tout2 - Tin2) Duty3 = F3*Cp3*(Tout3 - Tin3) Duty4 = F4*Cp4*abs(Tout4 - Tin4) Duty5 = F5*Cp5*abs(Tout5 - Tin5) Duty6 = F6*Cp6*abs(Tout6 - Tin6) heat = Duty1 + Duty2 + Duty3 cool = Duty4 + Duty5 + Duty6 steam = heat - cool printf("\n\nResult\n\n") printf("\n As a minimum %.0f Btu/h will have to be supplied by steam or another hot medium",steam)
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//This Source file is written by Nikita Rath (18BLC1131), VIT Chennai //Function for C-SCAN function [] = cscan(a,head,n) printf("Order of Track "); seek_count = 0; ihead = head; maximum = 4999; temp1 = 1; temp2 = 1; //Traversing through requests for i = 1:n //Request greater or less than head if(a(i)>=head) then queue1(temp1)=a(i); temp1 = temp1 + 1; else queue2(temp2)=a(i); temp2 = temp2 + 1; end end //Sort request greater than head (Ascending) for i = 1:temp1-2 for j = i+1:temp1-1 if (queue1(i) > queue1(j)) then temp = queue1(i); queue1(i) = queue1(j); queue1(j) = temp; end end end //Sort request less than head (Ascending) for i = 1:temp2-2 for j = i+1:temp2-1 if (queue2(i) > queue2(j)) then temp = queue2(i); queue2(i) = queue2(j); queue2(j) = temp; end end end //Request ordering begins i = 1; //Traversing to the right side for j = 1:temp1-1 i = i + 1; queue(i) = queue1(j); end //Reaches the maximum disk limit queue(i+1) = maximum; //Turns to the minimum disk limit queue(i+2) = 0; i = temp1+2; //Traversing to the left side for j = 1:temp2-1 i = i + 1; queue(i) = queue2(j); end queue(1)=head; //Traversing through final request order for j = 1:n+2 //Calculate the distance between requests distance = abs(queue(j+1)-queue(j)); //Increment seek_count with the distance seek_count = seek_count + distance; //Order of request execution printf(" T%d ",j); //printf(" - %d",queue(j+1)); end printf(" T%d ",j+1); printf("\nNo. of Cylinders"); printf(" %d ",ihead); //Order of request execution for j = 1:n+2 printf(" %4d ",queue(j+1)); end //Total distance printf("\n Total distance : %d", seek_count); endfunction
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// Example 14.9 // Calculating a Transfer Function // From figure 14.21 and ABCD parameters that we found in example 14.7 we have, // since V_1=V_s // Z_s=0; s=%s; disp("H(s)=I_2/V_1") disp("=> H(s)=-1/A*Z_L+B") A=1-2/s; B=-20/s; Z_L=2.5*s// Assume H_s=-1/(A*Z_L+B); P_s=(s^2-2*s-8); // denominator of H_s p=roots(P_s); disp(H_s,"Transfer function=") disp(P_s,"Characteristic polynomial=") disp(p,"Poles of transfer function=")
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//Determine the maximum value of transmitted wave clear clc; Z=350;//surge impedencr (ohms) C=3000*(10^-12);// earth capacitance(F) t=2*(10^-6); E=500; E1=2*E*(1-exp((-1*t/(Z*C)))); mprintf("the maximum value of transmitted voltage=%.0f kV \n",E1);
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//Example 8_2 <a> //determine the nyquist rate of x(t)=sinc(200*pi*t) //sinc(t)=cos(t)/t //cos3(t)=3/4[cos(200)+1/4cos(600)] clc; clear all; wp=200; F1=wp/2; Fs=2*F1; disp('Nyquist Rate='); disp(Fs);
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// Scilab Code Ex2.39:: Page-2.29 (2009) clc; clear; t = 0.75e-06; // Thickness of the glass plate, m mu = 1.5; // Refractive index of the glass plate lambda1 = 4000e-010; // First wavelength of visible range, cm lambda2 = 7000e-010; // Last wavelength of visible range, cm r = 0; // Angle of refraction for normal incidence, degrees n = zeros(2); // For bright fringe in reflected pattern, // 2*mu*t*cosd(r) = (2*n+1)*lambda/2, solving for n // For lambda1 n(1) = (4*mu*t*cosd(r)/lambda1-1)/2; // For lambda2 n(2) = (4*mu*t*cosd(r)/lambda2-1)/2; printf("\nFor n = %d and n = %d the light is strongly reflected.", n(1), ceil(n(2))); // Result // For n = 5 and n = 3 the light is strongly reflected.
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// This file is part of www.nand2tetris.org // and the book "The Elements of Computing Systems" // by Nisan and Schocken, MIT Press. // File name: projects/08/FunctionCalls/FibonacciElement/FibonacciElement.tst // FibonacciElement.asm is the result of translating both Main.vm and Sys.vm. load FibonacciElement.asm, output-file FibonacciElement.out, compare-to FibonacciElement.cmp, output-list RAM[0]%D1.6.1 RAM[261]%D1.6.1; repeat 6000 { ticktock; } output;
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clear all; clc; disp("Since the pressure changes are small compared with the barometric pressure,constant densities are assumed that is rho3 =rho2=rhoa where rhoa=pa/(RTa)") p_a=14.6 T_a=535 R=53.3 rho_a=(p_a*144)/(R*T_a)//144 is the conversion factor printf("rhoa= %0.4f lbm/ft^3",rho_a) A2=5*6.5 printf("\nA2 equals %g in^2",A2) disp("On converting A2=0.225ft^2") A3=%pi*6.065^2/4 printf("\nA3 equals %0.1f in^2",A3) disp("On converting A3=0.2007ft^2") disp("From (rho3*V3^2)/2=rhow*g*pv3,V3 can be calculated") rho_w=62.4 g=32.2 rho_3=0.0737 j=sqrt(2*rho_w*g/(rho_3*12)) printf("V3=%0.1f * pv3^0.5 ft/s",j) disp("The inlet flow rate can be calculated as Q=Q3=V3*A3=0.2007*60*V3=12.04*V3") disp("Also the dynamic pressure can be calculated as pv2=pv3*((A3/A2)^2)*(rho3/rho2)=(0.2/0.225)^2*pv3=0.79pv3") disp("The total pressure pt2=ps2+pv2") disp("To correct these data to a fixed speed of 3500rpm,the fan laws can be used as Qdash=Q*3500/N,pt2dash=pt2*((3500/N)^2) and Hdash=H*((3500/N)^3).") disp("The total efficiency can be calculated as ETAt=(rhow*g*pt2dash*Qdash)/(Hdash)") disp("On simplifying ETAt=Qdash*pt2dash/(6346*Hdash)") p_v3=[2.4 2.1 1.9 1.5 1.2 0.8 0.4]; p_s2=[2.5 4.2 6.0 6.8 7.6 8.5 9.5]; N=[3450 3520 3500 3420 3430 3500 3520]; H=[1.1 1.25 1.49 1.55 1.67 1.72 1.81]; V_3=zeros(1,length(p_v3)); Q=zeros(1,length(p_v3)); p_v2=zeros(1,length(p_v3)); p_t2=zeros(1,length(p_v3)); Qdash=zeros(1,length(p_v3)); tenpt2dash=zeros(1,length(p_v3)); tenHdash=zeros(1,length(p_v3)); eta_t=zeros(1,length(p_v3)); for i = 1: length(p_v3) V_3(i) = 67.4*sqrt(p_v3(i)); Q(i) = 12.04*V_3(i); p_v2(i) =0.79*p_v3(i); p_t2(i)= p_s2(i)+p_v2(i); Qdash(i)= Q(i)*(3500/(N(i))); tenpt2dash(i)= 10*p_t2(i)*((3500/N(i))^2); tenHdash(i)= 10*H(i)*((3500/(N(i)))^3); eta_t(i)= ((Qdash(i)*(tenpt2dash(i))/10)/(6346*(tenHdash(i))/10))*100; end disp("The table is in the order given in the book,that is pv3, ps2, N, H, V3, Q, pv2, pt2, Qdash, tenpt2dash, tenHdash and etat.") table=[p_v3' p_s2' N' H' V_3' Q' p_v2' p_t2' Qdash' tenpt2dash' tenHdash' eta_t' ]; disp(table) plot(Q,tenpt2dash,'o',Q,tenHdash,'d',Q,eta_t,'s') legend("tenpt2dash (inches of water)","tenHdash (hp)","eta_t (%)",-1) xlabel("Q(cfm)") ylabel("tenpt2dash (inches of water), tenHdash (hp) , eta_t (%)") set(gca(),"grid",[1 1])
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clc;clear; printf("\nBalanço de Massa\nMétodos Decomposição LU e Jacobi\n\n") /*ordem = input("Qual a ordem da matriz? ") printf("Preencha os valores da matriz dos coeficientes:\n") for i = 1:ordem printf("Linha %d\n", i) for j = 1:ordem printf("Coluna %d", j) A(i, j) = input("Valor: ") end end printf("Preencha os valores dos termos independentes:\n") for i = 1:ordem B(i) = input("Termo: ") end*/ n = 3; //A = [0.8 0.8 0.9; 0.05 0.9 0.9; 0.02 0.02 0.05]; //b = [500 300 200]; A = [-130 30 0; 90 -90 0; 40 60 -120]; b = [-200 0 -500] L = zeros(n,n); // Matriz triangular inferior com os coeficientes U = zeros(n,n); // Matriz que resta da eliminação de Gauss // Decomposição LU for j = 1:n L(j,j) = 1; // diagonal principal com 1 for i = 1:j soma = 0.0; for k=1:i-1 soma = soma + L(i,k)*U(k,j); end U(i,j) = A(i,j) - soma; // Cálculo da matriz U end for i = j+1:n soma = 0.0; for k=1:j-1 soma = soma + L(i,k)*U(k,j); end L(i,j) = (A(i,j)-soma)/U(j,j); // Cálculo da matriz L end end printf('Matriz L: \n') disp(L) printf('\nMatriz U: \n') disp(U) // resolve L*y = b: substituicao progressiva y = zeros(1,n); y(1) = b(1)/L(1,1); for i=2:n soma = 0.0; for j=1:i-1 soma = soma + L(i,j)*y(j); end y(i) = (b(i)-soma)/L(i,i); end // resolve U*x = y: substituicao regressiva x(n) = y(n)/U(n,n); for i=n-1:-1:1 soma = 0.0; for j=i+1:n soma = soma + U(i,j)*x(j); end x(i) = (y(i)-soma)/U(i,i); end printf('\nResultados pelo método da Decomposição LU: \n') disp(x) function [x, erro] = Jacobi(A, b, n, x) xAnt = x erro = 0.0 for i = 1:n soma = 0.0 for j = 1:n if (j <> i) then soma = soma + (A(i, j)*xAnt(j)) end end x(i) = (b(i) - soma)/A(i, i) if (abs(x(i) - xAnt(i)) > erro) erro = abs(x(i) - xAnt(i)) end end endfunction precisao = 0.001 // Critério de parada! maxIter = 100 // Critério de parada! k = 1 erro = 1000 x = [0; 0; 0] // Solução Inicial while (k < maxIter & erro > precisao) then [x, erro] = Jacobi(A, b, n, x) k = k + 1 end printf("\nResultados pelo método iterativo Jacobi: \n") disp(x)
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THE OPTIMIZATION ALGORITHM HAS CHANGED TO THE EM ALGORITHM. ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 0.409234D+00 2 -0.600470D-02 0.298930D-02 3 0.287096D-01 -0.628111D-03 0.265272D+00 4 -0.607017D-03 0.190080D-03 -0.230958D-02 0.214526D-02 5 0.628083D-03 0.884542D-05 -0.122089D-02 0.510410D-04 0.442288D-02 6 -0.145443D-02 0.895127D-04 0.877126D-04 0.428116D-04 0.163920D-03 7 -0.588537D-03 0.868556D-04 0.535821D-03 0.175638D-03 -0.331832D-03 8 0.177998D-02 -0.346083D-05 0.401361D-03 -0.243721D-04 -0.338197D-04 9 -0.427189D+00 0.630535D-02 -0.100268D+00 -0.338619D-02 0.208490D+00 10 -0.107615D+00 -0.609398D-02 0.396361D-01 0.541826D-02 0.235957D+00 11 -0.108569D+00 0.528872D-02 -0.102844D+00 0.301163D-02 0.187628D-02 12 0.515517D+00 -0.139242D-01 0.195080D+00 -0.125307D-02 0.291524D-01 13 0.142378D-01 0.392501D-02 0.665874D-01 0.112197D-01 -0.864610D-02 14 0.234568D-01 0.798119D-02 0.242018D+00 0.503095D-02 -0.216854D-01 15 -0.376483D+01 0.689497D-02 -0.108509D+01 0.165504D-01 -0.199778D+00 16 0.108678D-02 -0.119051D-01 0.599614D-02 -0.339623D-02 0.187032D-02 17 0.371953D-02 -0.654447D-04 0.922609D-03 -0.903262D-04 -0.106467D-02 18 -0.100580D+01 0.183313D-01 -0.548586D+00 -0.323700D-01 0.499094D-01 19 0.252045D-01 -0.271740D-02 0.962044D-01 -0.303538D-02 -0.662418D-02 20 -0.634381D+00 0.572207D-02 0.384391D+00 0.220868D-01 -0.580400D-01 21 -0.118683D-01 -0.415103D-02 -0.142371D+00 0.323475D-02 0.736462D-02 22 0.147408D-03 -0.826476D-04 0.327865D-02 -0.112125D-03 -0.711818D-04 23 -0.116610D-01 -0.637962D-03 0.185632D-01 0.246784D-02 0.181553D-02 24 0.173406D-02 -0.146482D-03 -0.692761D-03 0.169295D-03 -0.767654D-04 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 0.139913D-02 7 0.819704D-03 0.172728D-02 8 -0.934351D-05 0.172704D-03 0.248053D-02 9 -0.144991D-01 -0.147670D-01 -0.136919D-01 0.601987D+02 10 0.165782D-01 -0.110300D-01 0.580188D-03 0.116435D+02 0.248927D+02 11 0.257780D-01 -0.944556D-03 0.185696D-01 -0.143214D+01 0.242073D+01 12 0.748848D-03 0.230249D-01 0.797215D-01 0.155201D+01 0.804168D+00 13 0.449242D-01 0.557489D-01 0.529113D-02 -0.661527D+00 -0.164683D+01 14 0.376110D-02 -0.727169D-02 0.123021D+00 -0.217310D+01 -0.136696D-01 15 0.556154D-01 0.569981D-01 0.433392D-01 -0.172970D+02 -0.167897D+02 16 0.627201D-03 0.137263D-02 0.187336D-03 0.985575D+00 0.233543D+00 17 -0.403716D-03 -0.372085D-03 -0.397006D-03 -0.174649D+00 -0.680283D-01 18 -0.465741D-02 -0.433087D-01 0.127925D-01 0.510611D+01 0.455479D+00 19 -0.344571D-02 0.631692D-02 -0.195965D-02 -0.152819D+01 -0.189630D+00 20 0.273151D-01 0.228779D-01 -0.103798D+00 -0.114139D+01 0.542223D+00 21 0.409997D-02 -0.471264D-02 0.294213D-03 0.184018D+01 0.396190D+00 22 -0.495868D-03 -0.317721D-03 -0.282554D-04 -0.114655D-01 0.450706D-02 23 -0.775302D-03 -0.943663D-03 0.284390D-03 0.257330D+00 0.127546D+00 24 -0.157082D-03 -0.136226D-05 -0.298820D-03 -0.139140D-01 -0.189416D-01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 0.190230D+02 12 -0.374440D+01 0.426757D+02 13 -0.115546D+01 0.118805D+01 0.698389D+01 14 0.178748D+01 0.164060D+01 -0.349853D+00 0.203450D+02 15 0.729962D+01 0.409862D+01 0.300423D+01 0.238653D+01 0.467326D+03 16 0.177427D+00 -0.868150D-02 0.653839D-02 0.662157D-02 0.203901D+01 17 -0.388976D-01 -0.426458D-01 -0.299953D-01 -0.338259D-02 -0.193291D+01 18 -0.333866D+01 0.481958D+01 -0.174642D+01 -0.135240D+01 0.508643D+02 19 0.158113D+00 0.410370D+00 -0.894557D-01 -0.207075D+00 0.124866D+01 20 0.934194D+00 -0.173440D+02 0.205496D+01 -0.847035D+01 0.511139D+01 21 0.803031D-01 -0.302096D+00 0.136788D-02 0.200335D+00 -0.523808D+00 22 -0.114466D-01 -0.242058D-01 -0.254049D-01 0.748884D-02 -0.209540D+00 23 -0.323407D-01 0.163265D+00 -0.239018D-01 -0.144298D-01 -0.448544D+00 24 -0.165669D-01 -0.782930D-02 -0.355557D-02 -0.299096D-01 -0.100049D-01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 0.741613D+00 17 -0.571123D-01 0.221180D-01 18 0.392084D+00 -0.194825D+00 0.161416D+03 19 0.242272D+00 -0.136552D-01 -0.180812D+01 0.341557D+01 20 0.109388D-01 -0.147473D-02 -0.229870D+01 0.221891D+01 0.166329D+03 21 0.227399D-01 -0.105505D-01 0.324377D+01 -0.304210D+01 -0.296453D+01 22 -0.118746D-01 0.277990D-02 -0.714558D+00 0.822435D-02 0.509266D-01 23 0.252292D-01 -0.208582D-02 -0.399801D+00 -0.727744D-01 0.110386D+01 24 -0.386908D-02 0.666586D-03 0.652552D-01 -0.694525D-02 -0.757600D+00 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 0.379590D+01 22 -0.511015D-01 0.845816D-02 23 0.793710D-01 0.486220D-03 0.212960D+00 24 0.824856D-02 -0.674485D-03 -0.158574D-01 0.769294D-02 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 1.000 2 -0.172 1.000 3 0.087 -0.022 1.000 4 -0.020 0.075 -0.097 1.000 5 0.015 0.002 -0.036 0.017 1.000 6 -0.061 0.044 0.005 0.025 0.066 7 -0.022 0.038 0.025 0.091 -0.120 8 0.056 -0.001 0.016 -0.011 -0.010 9 -0.086 0.015 -0.025 -0.009 0.404 10 -0.034 -0.022 0.015 0.023 0.711 11 -0.039 0.022 -0.046 0.015 0.006 12 0.123 -0.039 0.058 -0.004 0.067 13 0.008 0.027 0.049 0.092 -0.049 14 0.008 0.032 0.104 0.024 -0.072 15 -0.272 0.006 -0.097 0.017 -0.139 16 0.002 -0.253 0.014 -0.085 0.033 17 0.039 -0.008 0.012 -0.013 -0.108 18 -0.124 0.026 -0.084 -0.055 0.059 19 0.021 -0.027 0.101 -0.035 -0.054 20 -0.077 0.008 0.058 0.037 -0.068 21 -0.010 -0.039 -0.142 0.036 0.057 22 0.003 -0.016 0.069 -0.026 -0.012 23 -0.040 -0.025 0.078 0.115 0.059 24 0.031 -0.031 -0.015 0.042 -0.013 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 1.000 7 0.527 1.000 8 -0.005 0.083 1.000 9 -0.050 -0.046 -0.035 1.000 10 0.089 -0.053 0.002 0.301 1.000 11 0.158 -0.005 0.085 -0.042 0.111 12 0.003 0.085 0.245 0.031 0.025 13 0.454 0.508 0.040 -0.032 -0.125 14 0.022 -0.039 0.548 -0.062 -0.001 15 0.069 0.063 0.040 -0.103 -0.156 16 0.019 0.038 0.004 0.148 0.054 17 -0.073 -0.060 -0.054 -0.151 -0.092 18 -0.010 -0.082 0.020 0.052 0.007 19 -0.050 0.082 -0.021 -0.107 -0.021 20 0.057 0.043 -0.162 -0.011 0.008 21 0.056 -0.058 0.003 0.122 0.041 22 -0.144 -0.083 -0.006 -0.016 0.010 23 -0.045 -0.049 0.012 0.072 0.055 24 -0.048 0.000 -0.068 -0.020 -0.043 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 1.000 12 -0.131 1.000 13 -0.100 0.069 1.000 14 0.091 0.056 -0.029 1.000 15 0.077 0.029 0.053 0.024 1.000 16 0.047 -0.002 0.003 0.002 0.110 17 -0.060 -0.044 -0.076 -0.005 -0.601 18 -0.060 0.058 -0.052 -0.024 0.185 19 0.020 0.034 -0.018 -0.025 0.031 20 0.017 -0.206 0.060 -0.146 0.018 21 0.009 -0.024 0.000 0.023 -0.012 22 -0.029 -0.040 -0.105 0.018 -0.105 23 -0.016 0.054 -0.020 -0.007 -0.045 24 -0.043 -0.014 -0.015 -0.076 -0.005 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 1.000 17 -0.446 1.000 18 0.036 -0.103 1.000 19 0.152 -0.050 -0.077 1.000 20 0.001 -0.001 -0.014 0.093 1.000 21 0.014 -0.036 0.131 -0.845 -0.118 22 -0.150 0.203 -0.612 0.048 0.043 23 0.063 -0.030 -0.068 -0.085 0.185 24 -0.051 0.051 0.059 -0.043 -0.670 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 1.000 22 -0.285 1.000 23 0.088 0.011 1.000 24 0.048 -0.084 -0.392 1.000
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x=[2.30256624769934; 2.29071803023829; 2.26283604900314; 2.35145015316178; 2.27686291358213; 2.29805616201205; 2.32805830340568; 2.30878734371402; 2.29343801980763; 2.23019030245799]; fs=4e6; t=(1/fs); [F,LT,UT]=falltime(x,fs); disp(F); disp(LT); disp(UT); //output // 0.0000002 // // 0.0000022 // // 0.0000020 //
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// Problem 5.1,Page no.121 clc;clear; close; b=100 //mm //width of timber joist d=200 //mm //depth of joist L=3 //m //Length of beam sigma=7 //KN/mm**2 //bending stress w_1=5 //KN/mm**2 //unit weight of timber //Calculations w=0.1*0.2*1*5*100 //N/m //self weight of the joist I_xx=1*12**-1*100*200**3 //mm**4 //M.I of section about N.A //M=W*L+w*L**2*2**-1 //Max Bending moment //Therefore,M=(3*W+450) //using the relation M*I**-1=sigma*y**-1,we get W=(((7*2*10**8)*(100*10**3*3)**-1)-450)*3**-1 //N //Max Load applied //Result printf("The Max value of Load applied is %.2f N",W)
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E = 29000; // Modulus of elasticity in ksi spl = 42 ; // Proportional limit in ksi L = 25 ; // Total length of coloum in ft n = 2.5 ; // factor of safety I1 = 98 ; // Moment of inertia on horizontal axis I2 = 21.7 ; // Moment of inertia on vertical axis A = 8.25 ; // Area of the cross section Pcr2 = (4*%pi^2*E*I2)/((L*12)^2) ; // Criticle load if column buckles in the plane of paper Pcr1 = (%pi^2*E*I1)/((L*12)^2) ; // Criticle load if column buckles in the plane of paper Pcr = min(Pcr1,Pcr2) ; // Minimum pressure would govern the design scr = Pcr/A ; // Criticle stress Pa = Pcr/n ; // Allowable load in k disp("k",Pa,"The allowable load is ")
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clear;lines(0); a=rand(3,3);p=poly([1,2,3],'s');l=list(1,'asdf',[1 2 3]); print(%io(2),a,p,l) write(%io(2),a)
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function [out]=erode(input_image ,kernel,actualkernel,anchorX,anchorY) input_image1=mattolist(input_image); a=opencv_erode(input_image1 ,kernel,actualkernel,anchorX,anchorY); dimension=size(a) for i = 1:dimension out(:,:,i)=a(i); end endfunction;
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function [x,y,typ] = mdaq_dio_get(job,arg1,arg2) dio_get_desc = ["This block reads MicroDAQ DIO input state."; ""; "DIO pin: 1...32"; ""; "Set block parameters:"]; x=[];y=[];typ=[]; select job case 'set' then x=arg1 model=arg1.model; graphics=arg1.graphics; exprs=graphics.exprs; while %t do try getversion('scilab'); [ok,dio_pin,exprs]=.. scicos_getvalue(dio_get_desc,.. ['DIO pin:'],.. list('vec',1),exprs) catch [ok,dio_pin,exprs]=.. scicos_getvalue(dio_get_desc,.. ['DIO pin:'],.. list('vec',1),exprs) end; if ~ok then break end if dio_pin > 32 | dio_pin < 1 then ok = %f; message("Wrong DIO pin selected, use value from 1 to 32!"); end if ok then [model,graphics,ok] = check_io(model,graphics, [], 1, 1, []); graphics.exprs = exprs; model.rpar = []; model.ipar = [dio_pin]; model.dstate = []; x.graphics = graphics; x.model = model; x.graphics.style=["mdaq_dio_get;blockWithLabel;verticalLabelPosition=center;displayedLabel=DIO%1$s;fontColor=#5f5f5f"] break end end case 'define' then dio_pin = 1; model=scicos_model() model.sim=list('mdaq_dio_get_sim',5) model.in =[] model.outtyp=1 model.out=1 model.out2=1 model.evtin=1 model.rpar=[]; model.ipar=[dio_pin] model.dstate=[]; model.blocktype='d' model.dep_ut=[%t %f] exprs=[sci2exp(dio_pin)] gr_i=['xstringb(orig(1),orig(2),[''DIO'' ; string(dio_pin)],sz(1),sz(2),''fill'');'] x=standard_define([4 3],model,exprs,gr_i) x.graphics.in_implicit=[]; x.graphics.exprs=exprs; x.graphics.style=["blockWithLabel;verticalLabelPosition=center;displayedLabel=DIO%1$s;fontColor=#5f5f5f"] end endfunction
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function [out]=convexhull(pstData,clkwise,returnpoints) out=opencv_convexhull(pstData,clkwise,returnpoints); endfunction;
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clear; clc; printf("\nEx1.7\n"); //page no.-9 //given rho=2700;.......//density of potassium bromide in kg/m^3 m=119;.........//molecular wt. n=4;...........//molecules per unit cell for F.C.C. N=6.02*10^26;...//avagadro no. M=(n*m)/N;..........//mass in each unit cell //as density=mass/volume, so volume is a^3 a=(M/rho)^(1/3)......//lattice constant in Angstrom printf("\nlattice constant is 6.64 angstrom\n");
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;Create new list /*declare json = vc go set json = '{"LISTREQUEST":{"patient_list_id":0.0,"name":"Test List 1","description":"","patient_list_type_cd":2177315703.0,"\ owner_prsnl_id":18811197.0,"arguments":[{"argument_name":"ACMPRSNLGROUPS","argument_value":"","parent_entity_id":20705740.0,"\ parent_entity_name":"PRSNL_GROUP"}{"argument_name":"PPRCODES","argument_value":"","parent_entity_id":1115.0,"parent_entity_name\ ":"PERSON_PRSNL_RELTN"}{"argument_name":"GENDER","argument_value":"","parent_entity_id":363.0,"parent_entity_name":"CODE_VALUE"}{"argument_name":"AGEMIN","argument_value":"18","parent_entity_id":0.0,"parent_entity_name":""}]}}' go execute mp_dcp_upd_patient_list "MINE", json go */ ; Update existing list declare json = vc go set json = '{"LISTREQUEST":{"patient_list_id":4386056.0,"name":"Test List 1","description":"","patient_list_type_cd":2177315703.0,"\ owner_prsnl_id":18811197.0,"arguments":[{"argument_name":"ACMPRSNLGROUPS","argument_value":"","parent_entity_id":20523635.0,"\ parent_entity_name":"PRSNL_GROUP"}{"argument_name":"PPRCODES","argument_value":"","parent_entity_id":1115.0,"parent_entity_name\ ":"PERSON_PRSNL_RELTN"}{"argument_name":"GENDER","argument_value":"","parent_entity_id":363.0,"parent_entity_name":"CODE_VALUE"}{"argument_name":"AGEMIN","argument_value":"18","parent_entity_id":0.0,"parent_entity_name":""}]}}' go execute mp_dcp_upd_patient_list "MINE", json go /* free record request go record request ( 1 owner_prsnl_id = f8 ) go set request->owner_prsnl_id = 18811197.0 go mp_dcp_retrieve_patient_lists go*/
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load HackComputer.hdl; output-file TestCaseIntegerArithmetic.out; //Loading program to instruction memory ROM32K load TestCaseIntegerArithmetic.hack; //Expression being evaluated is d=a+b-c with - //a=100 //b=50 //c=10 //Value of d is stored at address 19 output-list RAM64[19]%D1.3.1; set reset 1; tick,tock; set reset 0; output; repeat 20 { tick,tock; } output;
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//------------------------------------------------------------------------------ // FEDERAL UNIVERSITY OF UBERLANDIA // Faculty of Electrical Engineering // Biomedical Engineering Lab // Uberlandia, Brazil //------------------------------------------------------------------------------ // Author: Andrei Nakagawa, MSc // Contact: andrei.ufu@gmail.com //------------------------------------------------------------------------------ //------------------------------------------------------------------------------ // Description: Combining a sensorimotor transformation // function with optimal feedback control to explain online // corrections to visual perturbations. //------------------------------------------------------------------------------ // Algorithm: Before running the simulation, find the optimal // time-varying feedback gains according to the task goal. // Using a delayed visual feedback, whenever target location // in motor space changes, the gains should be recomputed // given the amount of time left to complete the reach and // applied to the simulation. Corrections are only necessary // if cursor is not inside the target (to avoid error oscillations). // Delayed feedback and reaction times will be combined // (see Li and Todorov, 2007) with a constant value (200 ms). //------------------------------------------------------------------------------ //Kalman filter function [xplus,pplus] = kalman(F,G,H,Q,R,xk,pk,yk,uk) //time update (prediction) xminus = F*xk + G*uk; pminus = F*pk*F' + G*Q*G'; //measurement update (correction) measureError = yk - (H*xminus); kalmanGain = pminus*H' * inv(H*pminus*H' + R); //state estimate xplus = xminus + kalmanGain * measureError; //covariance pplus = pminus - kalmanGain*H*pminus; endfunction //------------------------------------------------------------------------------ //Optimal time-varying feedback gains function Kopt = computeGain(Ad,Bd,Qd,Rd,S0,time) Kopt = []; for k=1:length(time) //Calculating the time-varying gain K = inv(Bd'*S0*Bd + Rd)*(Bd'*S0*Ad); //New riccati solution S0 = Ad'*S0*Ad - Ad'*S0*Bd*((Rd + Bd'*S0*Bd)^-1)*Bd'*S0*Ad + Qd; //Stores the gain Kopt = [Kopt K]; end endfunction //------------------------------------------------------------------------------ //States //Position and velocity in X //Position and velocity in Y //Inputs //Force in X and Y function [Ac,Bc,Cc] = pointMassModel(m) Ac = [0 1 0 0; 0 0 0 0; 0 0 0 1; 0 0 0 0]; Bc = [0 0; 1/m 0; 0 0; 0 1/m]; Cc = eye(size(Ac,1),size(Ac,2)) endfunction //------------------------------------------------------------------------------ //Function that specifies the task goal in motor space according to visual space function mtarget=smTransf(vcursor, vtarget, mhand) //movement vector in visual space vvector = vtarget - vcursor; //magnitude magnitude = norm(vvector); //direction if(vvector(1) == 0) direction = (90*%pi)/180; else direction = abs(atan(vvector(2)/vvector(1))); end //estimating target location in motor space mtarget = zeros(1,2); //x-axis mtarget(1) = mhand(1) + (sign(vvector(1)) * magnitude * cos(direction)); //y-axis mtarget(2) = mhand(2) + (sign(vvector(2)) * magnitude * sin(direction)); endfunction //------------------------------------------------------------------------------ [A,B,C] = pointMassModel(1); //------------------------------------------------------------------------------ //Simulation parameters t0=0; tf=6; dt = 0.01; t = t0:dt:tf; //Continuous-time system contSys = syslin('c',A,B,C); //Discrete-time system discSys = dscr(contSys,dt); //------------------------------------------------------------------------------ //------------------------------------------------------------------------------ //Weight matrices Qd=diag([0.1,0.1,0.1,0.1]); Rd=diag([0.001,0.001]); //Discrete riccati Ad = discSys(2); //A Bd = discSys(3); //B //------------------------------------------------------------------------------ //------------------------------------------------------------------------------ //Differential Riccati Equation - Discrete-time //Calculating the solution to riccati for each instant in time //and then finding the time-varying gain for each time step //------------------------------------------------------------------------------ Sdisc = []; Kdisc = []; S0 = diag([50,0,50,0]); //Estimate for the Riccati matrix for k=1:length(t) //Calculating the time-varying gain K = inv(Bd'*S0*Bd + Rd)*(Bd'*S0*Ad); //New riccati solution S0 = Ad'*S0*Ad - Ad'*S0*Bd*((Rd + Bd'*S0*Bd)^-1)*Bd'*S0*Ad + Qd; //Stores the riccati solution Sdisc = [Sdisc S0]; //Stores the gain Kdisc = [Kdisc K]; end //------------------------------------------------------------------------------ cont = 1; //Desired setpoints or reference trajectory xd = [0;0;5;0]; vtarget = [xd(1),xd(3)]; xint = []; //stores all the states during integration uint = []; //stores all the inputs during integration costQ = [0]; //cost of states costR = [0]; //cost of control x0 = [0;0;0;0]; //temporary variable for storing states u0 = [0;0]; x = x0; xint = [xint x0]; uint = [uint u0]; //Perturbation //Rotation matrix perturbation = 30; //degrees Ck = [cos((perturbation*%pi)/180) 0 -sin((perturbation*%pi)/180) 0; sin((perturbation*%pi)/180) 0 cos((perturbation*%pi)/180) 0]; yint = [x0]; yaux = x0; //Delayed feedback and reaction time delay = 0.200; //Every 200 ms, target in motor space will be updated t0 = t(1); //first time-step //Target jump targetJump = 0; //------------------------------------------------------------------------------ for k=1:length(t)-1 //Calculating the input (force) u = -Kdisc(:,cont:cont+3) * (x-xd); //Calculating the new states x = Ad*x + Bd*u; //perturbation y = Ck*x; y = [y(1);0;y(2);0]; aux = [yint y]; yvx = diff(yint(1,:)); if yvx == [] then yvx = 0; end yvy = diff(yint(3,:)); if yvy == [] then yvy = 0; end yaux(1) = y(1); yaux(2) = yvx($); yaux(3) = y(3); yaux(4) = yvy($); yint = [yint yaux]; //Checks if visual feedback should be updated if(t(k) - t0 >= delay) t0 = t(k); vcursor = [yint(1,k),yint(3,k)]; //vtarget = [xd(1), xd(3)]; mhand = [xint(1,k),xint(3,k)]; motorTarget = smTransf(vcursor,vtarget,mhand); if(motorTarget ~= [xd(1),xd(3)]) xd(1) = motorTarget(1); xd(3) = motorTarget(2); newTime = t(k:$); Kdisc = computeGain(Ad,Bd,Qd,Rd,S0,newTime); cont = 1; disp('updated'); end end //Storing the new states xint = [xint x]; //Storing the new inputs uint = [uint u]; //Stores the cost in this step costQ = [costQ (x-xd)'*Qd*(x-xd)]; //Stores the cost in this step costR = [costR u'*Rd*u]; //Increments the counter to loop through the gain matrix cont = cont + 4; end //------------------------------------------------------------------------------ //------------------------------------------------------------------------------ figure(); plot(xint(1,:),xint(3,:),'k'); plot(yint(1,:),yint(3,:),'b'); plot(xint(1,$),xint(3,$),'k.'); plot(vtarget(1),vtarget(2),'r.'); plot(x0(1),x0(3),'b.'); xlabel('x-axis','fontsize',4); ylabel('y-axis','fontsize',4); legend('Hand movement', 'Cursor motion'); a=get("current_axes"); a.font_size=3; ax=gca(); ax.data_bounds=[-3 -1; 3 6]; ax.labels_font_size=3; figure(); plot(t,xint(2,:),'r'); plot(t,xint(4,:),'g'); plot(t,uint(1,:),'k'); plot(t,uint(2,:),'k'); figure(); plot(t,costQ,'r'); plot(t,costR,'b'); title("Cost"); //------------------------------------------------------------------------------
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clear clc disp("2*x1*x2+2*x1*x3-2*x2*x3 ") disp("The matrix of the given quadratic form is ") A=[0 1 1;1 0 -1;1 -1 0] disp("let R represents the matrix of transformation and P represents a diagonal matrix whose values are the eigen values of A.then ") [R P]=spec(A) disp("so,canonical form is -2*x^2+y^2+z^2")
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clear;lines(0); genlib('auto1','SCI/macros/auto') disp(auto1)
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clc //Initialization of variables m=2 M=28 M2=32 PN=300 //psia Pt=400 //psia //calculations nN=m/M PO=Pt-PN nO=nN*PO/PN mO=M2*nO //results printf("Mass of oxygen added = %.3f lbm",mO)
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// To convert flux density to different units // Modern Electronic Instrumentation And Measurement Techniques // By Albert D. Helfrick, William D. Cooper // First Edition Second Impression, 2009 // Dorling Kindersly Pvt. Ltd. India // Example 2-2 in Page 29 clear; clc; close; // Given data B_cm = 20; // flux density in maxwell/sq.cm //Calculations B_in = B_cm *2.54^2; // converting to lines/sq.inch printf("The flux density in lines/sq.in = %d lines/(in^2)",B_in); //Result // The flux density in lines/sq.in = 129 lines/(in^2)
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%lsslp.sci
function [sr]=%lsslp(s,p) //sr=%lsslp(s,p) <=> sr=s\p // p polynomial matrix // s syslin list //! //origine S Steer INRIA 1992 sr=s\tlist(['lss','A','B','C','D','X0','dt'],[],[],[],p,[],[])
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Example8_32.sce
//Example 8.32 clc disp("Cascading four 74161 (each 4-bit) counters we get 16 (4 x 4) bit counter as shown in fig 8.63.") disp("Therefore, we get 2^16 = 65,536 modulus counter") disp("However, we require divide-by-40,000 counter. The difference between 65,536 and 40,000 is 25,536, which is the number of states those must be skipped from the full modulus sequence. This can be achieved by presetting the counting from 25,536 upto 65,536 on each ful cycle. Therefore, each full cycle of the counter consists of 40,000 states.")
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serveur3_histo.sce
clf; clear; clc; load('C:\Users\tangu\OneDrive\Documents\GitHub\Modelisation\TD4\NetworkData.sod') // Extraction des temps de service index_bool = ( data(:, 3) == 3 ) tabS3 = data(index_bool, :) t_s3 = tabS3(1:$,4); deciles=perctl(t_s3,10:10:90); for i=2:10 ClassesDeciles(i)=deciles(i-1) end ClassesDeciles(1)=min(t_s3) ClassesDeciles(11)=max(t_s3) histplot(ClassesDeciles,t_s3,style=2) legend("Histogramme d isofréquence du serveur 3") // Définition des paramètres d'affichages a=gca(); a.x_location = "origin"; a.grid=[5,5];
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example22_1.sce
clc // Given that d = 12e-6 // diameter in m d_ = 90e-9 // diameter of nanoparticle in m // Sample Problem 1 on page no. 22.13 printf("\n # PROBLEM 1 # \n") r = d / 2 r_ = d_ / 2 k = r / 3 k_ = r_ / 3 R = k_ / k printf("\n The ratio of the value of Nb/Ns of spherical particle and nanoparticle = %e .",R)
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example12_4.sce
disp('chapter 12 ex12.4') disp('given') disp("output =10V to 15V") Vomax=15 disp("max load current=4000mA") Il=.4 disp("Vsmin=Vomax+3 V") Vsmin=Vomax+3 disp('volts',Vsmin) disp("allowing Vrs=3V(p to p)") Vrs=3 disp("Vs=Vsmin+Vrs/2") Vs=Vsmin+Vrs/2 disp('volts',Vs) disp("ZENER CIRCUIT") disp("let Vz=Vo/2") Vz=Vomax/2 disp('volts',Vz) disp("Iz=20mA") Iz=.02 disp("R1=(Vo-Vz)/Iz") R1=(Vomax-Vz)/Iz disp('ohms',R1) disp("R1=330 ohm std value") R1=390 disp("POTENTIAL DIVIDER") disp("let I2&gt;&gt;Ibmax I2=50uA Vomin=10") I2=50*10^(-6) Vomin=10 disp("R2=(Vomin-Vz)/I2") Vz=7.5 R2=(Vomin-Vz)/I2 disp('ohms',R2) disp("R2=47kohm std value") R2=47000 disp("I2=(Vomin-Vz)/R2") I2=(Vomin-Vz)/R2 disp('amperes',I2) disp("R34=R3+R4=Vz/Iz") R34=Vz/I2 disp('ohms',R34) disp("when Vo is at its max,moving contact is at bottom of R4") disp("I2=Vomax/(R2+R34)") I2=Vomax/(R2+R34) disp('amperes',I2) disp("R3=Vz/Iz") R3=Vz/I2 disp('ohms',R3) disp("use 100 k ohm std value") R3=100000 disp("R4=(R3+R4)-R3") R4=R34-R3 disp('ohms',R4) disp("use 50 k ohm std value") disp("CAPACITOR") disp("select C1=100uF") C1=100*10^(-6) disp("Q1 specification") disp("Vcemax=Vsmax=Vs+Vrs/2") Vcemax=Vs+Vrs/2 disp('volts',Vcemax) Ie=Il disp("P=Vce*Il=(Vs-Vomin)*Il") P=(Vs-Vomin)*Il disp('watts',P) disp("A 2N3055 is a suitable device") disp("Q2 specification") disp("Vcemax=Vsmax=Vs+Vrs/2") Vcemax=Vs+Vrs/2 disp('volts',Vcemax) disp("Ie=Il/hFE1 ,hFE1=20 for Q1") hFE1=20 Ie=Il/hFE1 disp('amperes',Ie) disp("P=Vce*Il=(Vs-Vomin)*Il") P=(Vs-Vomin)*Il disp('watts',P) disp("A 2N3904 is a suitable device") disp("R5 Calculation") disp("let Ie2min=0.5mA,Vbe1=0.7") Ie2min=0.5*10^(-3) Vbe1=0.7 disp("R5=(Vomin+Vbe1)/Ie2min") R5=(Vomin+Vbe1)/Ie2min disp('ohms',R5) disp("R5=18kohm std value") disp("OPERATIONAL AMPLIFIER") disp("because I2 is sselected for bipolar opamp either a bipolar or BIFEt opamp can be used") disp("supply voltage Vs=19.5V") Vs=19.5 disp("Input supply voltage range=Vs/2-Vz") ipvoltage=(Vs/2)-Vz disp('volts',ipvoltage)
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unattended_rivalry.sce
scenario = "Unattended Rivalry"; response_matching = simple_matching; active_buttons = 2; button_codes = 1,2; default_font = "Calibri"; default_font_size = 18; default_text_color = 100,100,100; default_clear_active_stimuli = false; response_logging = log_all; write_codes = true; pulse_width = 6; response_port_output=true; begin; #intro text text {caption = "Click the mouse when you see a letter from two positions back repeated. Focus only on the letters at the center of the screen and do not move during the experiment.\n When instructed, click to continue.\n Good luck!"; max_text_width = 400; } intro_text; #fixation text text{caption = "When you have focused on the letter above, click to begin the experiment"; max_text_width = 400;} fixation_text; #feedback text text {caption = " ";} feedback_text; #trial beginning text text {caption = "Trial beginning...";} next_trial; #end of trial text text {caption = "Finished!\nPlease wait for further instructions.";} end_text; #letter array array { bitmap {filename = "\letters\\A.png";} A; bitmap {filename = "\letters\\B.png";} B; bitmap {filename = "\letters\\C.png";} C; bitmap {filename = "\letters\\D.png";} D; bitmap {filename = "\letters\\E.png";} E; bitmap {filename = "\letters\\F.png";} F; bitmap {filename = "\letters\\G.png";} G; bitmap {filename = "\letters\\H.png";} H; bitmap {filename = "\letters\\I.png";} I; bitmap {filename = "\letters\\J.png";} J; bitmap {filename = "\letters\\K.png";} K; bitmap {filename = "\letters\\L.png";} L; bitmap {filename = "\letters\\M.png";} M; bitmap {filename = "\letters\\N.png";} N; bitmap {filename = "\letters\\O.png";} O; bitmap {filename = "\letters\\P.png";} P; bitmap {filename = "\letters\\Q.png";} Q; bitmap {filename = "\letters\\R.png";} R; bitmap {filename = "\letters\\S.png";} S; bitmap {filename = "\letters\\T.png";} T; bitmap {filename = "\letters\\U.png";} U; bitmap {filename = "\letters\\V.png";} V; bitmap {filename = "\letters\\W.png";} W; bitmap {filename = "\letters\\X.png";} X; bitmap {filename = "\letters\\Y.png";} Y; bitmap {filename = "\letters\\Z.png";} Z; } letters; #template bitmap {filename = "\letters\\template.png";} template; #frequency-tagged houses array { LOOP $i 10; bitmap {filename = "house_tag_$i.png";} "house$i"; ENDLOOP; } houses; #frequency-tagged faces array { LOOP $i 10; bitmap {filename = "face_tag_$i.png";} "face$i"; ENDLOOP; } faces; #frequency-tagged boxes array { LOOP $i 10; bitmap {filename = "box_tag_$i.png";} "box$i"; ENDLOOP; } boxes; #frequency-tagged helmets array { LOOP $i 10; bitmap {filename = "helmet_tag_$i.png";} "helmet$i"; ENDLOOP; } helmets; #helmet-face transitions array { LOOP $i 49; bitmap {filename = "helmet_face_$i.png";} "helmet-face$i"; ENDLOOP; } helmets_faces; array { LOOP $i 49; bitmap {filename = "helmet_face_alt_$i.png";} "helmet-face-alt$i"; ENDLOOP; } helmets_faces_alt; array { LOOP $i 49; bitmap {filename = "helmet_face_dif_$i.png";} "helmet-face-dif$i"; ENDLOOP; } helmets_faces_dif; #box-house transitions array { LOOP $i 49; bitmap {filename = "box_house_$i.png";} "box-house$i"; ENDLOOP; } boxes_houses; array { LOOP $i 49; bitmap {filename = "box_house_alt_$i.png";} "box-house-alt$i"; ENDLOOP; } boxes_houses_alt; array { LOOP $i 49; bitmap {filename = "box_house_dif_$i.png";} "box-house-dif$i"; ENDLOOP; } boxes_houses_dif; #placeholder image bitmap {filename = "lines_small.png";} placeholder; #fixation mark picture { bitmap {filename = "lines.png";}; x = 0; y = 0; bitmap A; x = -250; y = 0; bitmap A; x = 250; y = 0; text fixation_text; x = -250; y = -250; text fixation_text; x = 250; y = -250; } fixation_pic; #trial picture picture { bitmap {filename = "lines.png";}; x = 0; y = 0; bitmap placeholder; x = -250; y = 75; bitmap template; x = -250; y = 0; bitmap placeholder; x = 250; y = 75; bitmap template; x = 250; y = 0; } trial_pic; #wait trial trial { trial_duration = forever; trial_type = first_response; picture { text intro_text; x = -250; y = 0; text intro_text; x = 250; y = 0; }; } wait_trial; #trial beginning trial { trial_duration = 1000; trial_type =fixed; picture { text next_trial; x = -250; y = 0; text next_trial; x = 250; y = 0; }; } begin_trial; #fixation trial trial { trial_duration = forever; trial_type = first_response; picture fixation_pic; } fixation_trial; #stimulus trial; variable duration trial { trial_type = fixed; terminator_button = 2; stimulus_event { picture trial_pic; stimulus_time_in = 100; stimulus_time_out = 2000; } stim_event; } stim_trial; #feedback trial trial { trial_duration = forever; trial_type = first_response; picture { text feedback_text; x = -250; y = 0; text feedback_text; x = 250; y = 0; }feedback_pic; } feedback_trial; #end trial trial { trial_duration = forever; trial_type = first_response; stimulus_event{ picture { text end_text; x = -250; y = 0; text end_text; x = 250; y = 0; } end_pic; port_code = 255; } end_event; } end_trial; ############################################################################## begin_pcl; ############################################################################## #variables double scenario_time = 2.4e5; #length of the scenario in ms: 6e5=10mins;4.8e5=8mins;1.2e5=2mins double before_trans_time = 5.8e4;#time before switching from the non-target images in ms double target_prop = 0.2; #proportion of targets in a trial int n_back = 2; #n-back value ############################################################################## #trial durations int b_dur = 78; #duration of frequency-tagged image b, 12.5 Hz int c_dur = 58; #duration of frequency-tagged image c, 16.67 Hz int gcd = 18; #greatest common demominator of b_dur and c_dur #letter timing int letter_dur; int total_dur; #stimuli and image parameters int num_targs = 0; #number of targets presented so far int num_non_targs = 0; #number of non-targets presented so far array<int> is_target[0]; #array of target positions array<int> last_left_image[1] = {1}; array<int> last_right_image[1] = {1}; int left_image = 1; int right_image = 1; int trans_length = 49; #modulo parameters int left_mod = 3; int right_mod = 4; int trans_mod = 5; #last n stimuli array<int> last_n_letters[0]; loop int x = 1 until x > n_back begin last_n_letters.add(1); x = x + 1; end; #event codes int box_img = 1; int helmet = 2; int house = 3; int face = 4; #face-house transitions array<bitmap> box_house_trans[0][0]; box_house_trans.add(boxes_houses); box_house_trans.add(boxes_houses_alt); box_house_trans.add(boxes_houses_dif); #house-face transitions array<bitmap> helmet_face_trans[0][0]; helmet_face_trans.add(helmets_faces); helmet_face_trans.add(helmets_faces_alt); helmet_face_trans.add(helmets_faces_dif); int trans_set_count = box_house_trans.count(); #for feedback int correct_count = 0; int false_alarm_count = 0; int total_hits = 0; int total_fas = 0; #width of stimuli from file input_file get_width = new input_file; get_width.open("\logfiles\\width_setting.txt"); int width = get_width.get_int(); get_width.close(); fixation_pic.set_part_x(2, -width); fixation_pic.set_part_x(3, width); #fixation_pic.set_part_x(4, -width); #fixation_pic.set_part_x(5, width); trial_pic.set_part_x(2, -width); trial_pic.set_part_x(3, -width); trial_pic.set_part_x(4, width); trial_pic.set_part_x(5, width); ############################################################################## #subroutines ############################################################################## #generates random stimulus array index sub int random_exclude(int first, int last, array<int> exclude[n_back]) begin int rval = random(first, last - 1); loop int i = 1; until i > exclude.count() begin if rval == exclude[i] then rval = random_exclude(first, last, exclude); i = 1; else i = i + 1; end; end; return rval end; #generates an array of target positions sub array<int, 1> is_target(int letter_count) begin array<int> targs[0]; loop int i = 1 until i > letter_count begin if i <= int(floor(double(letter_count) * target_prop)) then targs.add(1); else targs.add(0); end; i = i + 1; end; loop int i = 1 until i > n_back begin if targs[i] == 0 then i = i + 1; else targs.shuffle(); i = 1; end; end; return targs end; #runs a trial sub do_run(int run, int before_trans, int duration, array<bitmap> non_targs[2][10], array<bitmap> transitions[2][trans_set_count][trans_length], array<bitmap> targets[2][10]) begin array<bitmap> images[2][10] = non_targs; bool transitioned = false; int trans_index = 1; int trans_set = 1; int letter_index = 1; int letter_counter = 1; loop int k = 1; until k > duration begin if k > before_trans then images = targets; end; if transitioned || (k <= before_trans) then if mod(k,left_mod) == 0 then left_image = random_exclude(1, 10, last_left_image); trial_pic.set_part(2, images[1][left_image]); last_left_image[1] = left_image; end; if mod(k,right_mod) == 0 then right_image = random_exclude(1, 10, last_right_image); trial_pic.set_part(4, images[2][right_image]); last_right_image[1] = right_image; end; else if mod(k,trans_mod) == 0 then trial_pic.set_part(2, transitions[1][trans_set][trans_index]); trial_pic.set_part(4, transitions[2][trans_set][trans_index]); trans_set = trans_set + 1; if mod(trans_set,4) == 0 then trans_index = trans_index + 1; trans_set = 1; end; if trans_index > trans_length then transitioned = true; end; end; end; #letters if mod(k,total_dur) == 0 then if is_target[letter_counter] == 1 then letter_index = last_n_letters[n_back]; num_targs = num_targs + 1; stim_event.set_target_button(1); stim_event.set_stimulus_time_out(2000); else letter_index = random_exclude(1, 26, last_n_letters); num_non_targs = num_non_targs + 1; stim_event.set_target_button(0); stim_event.set_response_active(true); stim_event.set_stimulus_time_out(total_dur*gcd); end; trial_pic.set_part(3, letters[letter_index]); trial_pic.set_part(5, letters[letter_index]); letter_counter = letter_counter + 1; loop int j = last_n_letters.count(); until j == 1 begin last_n_letters[j] = last_n_letters[j-1]; j = j - 1; end; last_n_letters[1] = letter_index; end; if mod(k,total_dur) == letter_dur then trial_pic.set_part(3, template); trial_pic.set_part(5, template); end; stim_trial.present(); if mod(k,total_dur) == 0 then stim_event.set_target_button(0); stim_event.set_event_code(""); stim_event.set_port_code(0); end; k = k + 1; end; end; #displays results sub display_results begin correct_count = response_manager.total_hits() - total_hits; false_alarm_count = response_manager.total_false_alarms() - total_fas; double performance = double(correct_count - false_alarm_count) / double(num_targs); string cap = "Results:"; cap = cap + "\nNumber of Targets: " + string(num_targs); cap = cap + "\nHits: " + string(correct_count); cap = cap + "\nFalse Alarms: " + string(false_alarm_count); cap = cap + "\nPercentage Correct: " + printf(performance * 100.0, "%.1f") + "%"; cap = cap + "\n\nPlease call the experimenter in.\nWhen instructed, press the mouse to continue."; feedback_text.set_caption(cap); feedback_text.redraw(); feedback_trial.present(); total_hits = total_hits + correct_count; total_fas = total_fas + false_alarm_count; num_targs = 0; num_non_targs = 0; end; #main sub main begin input_file in = new input_file; in.open("\logfiles\\difficulty.txt"); letter_dur = int(in.get_line()) / gcd; total_dur = int(in.get_line()) / gcd; in.close(); wait_trial.present(); int duration = int(scenario_time) / gcd; int before_trans = int(before_trans_time) / gcd; array<bitmap> non_targs[0][0]; array<bitmap> targs[0][0]; array<bitmap> trans[0][0][0]; is_target.resize(1 + (duration / total_dur)); is_target = is_target(is_target.count()); loop int i = 1 until i > 2 begin begin_trial.present(); fixation_trial.present(); if i == 1 then non_targs.add(boxes); non_targs.add(helmets); trans.add(box_house_trans); trans.add(helmet_face_trans); targs.add(houses); targs.add(faces); else non_targs.add(helmets); non_targs.add(boxes); trans.add(helmet_face_trans); trans.add(box_house_trans); targs.add(faces); targs.add(houses); end; do_run(i, before_trans, duration, non_targs, trans, targs); display_results(); i = i + 1; end; end_trial.present(); end; main();
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7_3.sce
//To Calculate the Voltage Drop and Verify The Cable Selected //Page 329 clc; clear; pf=0.9; //Power Factor Vb=120; //Base Voltage //From The Tables r=0.334; //Resistance per thousand feet x=0.0299; //Reactance per thousand feet K1=0.02613; //Voltage Drop //Assumed Cable I=100; //Secodary line Current Ls=100; //Length of Secondary line in feet R=r*Ls/1000; // Resistance Value for a 100 feet Line X=x*Ls/1000; // Reactance Value for a 100 feet Line VD=I*((R*pf)+(X*sind(acosd(pf)))); //Voltage Drop VDpu=VD/Vb; //Per unit value printf('\nThe Cable Selected is of 100 feet, carrying 100A and cable size #2 AWG\n') printf('The Voltage drop for the above cable is %g pu V\n',VDpu) printf('The Above Value is Close to the Value(%g pu V) in the table given.\n',K1)
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//Example 1.5 (c) //To Determine Whether Given Signal is Periodic or not clear; clc ; close ; t=0:0.01:10; x1=cos(2*%pi*t/3); subplot(1,2,1); plot(t,x1); xlabel('t'); ylabel('x(t)'); title('CONTINUOUS TIME PLOT'); n=0:0.2:10; x2=cos(2*%pi*n/3); subplot(1,2,2); plot2d3(n,x2); xlabel('n'); ylabel('x(n)'); title('DISCRETE TIME PLOT'); //Hence Given Signal is Periodic with N=3
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example5_9.sce
//example 5.9 //calculate input h.p of pump clc; //given A=20; //area of field H=129; //level to the highest land h1=120.2; //water level in well during discharge Du=800; //duty for rise; eita=0.6; //efficiency of the pump Q=A/Du; w=Q*1000; lift=H-h1; //design lift is taken as 9m wd=w*9; o=wd/75; i=o/eita; mprintf("Input h.p of pump=%i h.p",i);
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clear; //clc(); // Example 11.8 // Page: 287 printf("Example-11.8 Page no.-287\n\n"); //***Data***// T_i = 50;//[F] Initial temperature of the system T_f = 20;//[F] Final temperature of the system M_gas = 115;//[g/mol] Molecular weight of gasoline at room temperature M_water = 18;//[g/mol] Molecular weight of water at the room temperaature d = 720;//[g/L] density of gasoline at the room temperature // From Figure 11.10 ( page 288 ), solubility of the water in gasoline ( similar to solubility of water in cyclohexane ) at 50 deg F is given as s_50 = 0.00026;//[mol fraction] // And linearly extraploting the cyclohexane curve in figure 11.10 to 20 deg F, we get the solubility of water at 20deg F as s_20 = 0.0001;//[mol fraction] // So, rejected water is s_rej = s_50 - s_20;// mol of water per mole of gasoline // In terms of weight, rejected water will be w = (s_rej*d*M_water)/M_gas;//[g water/L gasoline] printf(" The amount of water that will come out of the solution in the gasoline will be %f g water/L gasoline\n",w); printf(" At 20 deg F we would expect this water to become solid ice, forming a piece large enough to plug the fuel line of a parked auto.");
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clear; clc; vf=1; r=1250e3; V=600; x1=.5; x2=.5; x3=.02; ia2=vf/(x1+x2+x3); ia=3*ia2; ia1=ia2; ia0=ia1; iab=r/(sqrt(3)*V); iab=round(iab/10)*10; ia=round(ia*100)/100; If=ia*iab;//the difference in result is due to erroneous calculation in textbook. printf("fault current If=%fA",If); disp("the difference in result is due to erroneous calculation in textbook.")
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PL/SQL Developer Test script 3.0 21 -- Created on 08.09.2014 by ZHURAVOV_VB declare -- Local variables here a anydata; o xxdoo_cntr_contractor_typ; c xxdoo_cntr_contractors_typ; -- procedure show(a anydata) is l_type_code pls_integer; l_type anytype; begin l_type_code := a.GetType(l_type); dbms_output.put_line(l_type_code); end; begin -- Test statements here show(anydata.ConvertObject(o)); show(anydata.ConvertCollection(c)); end; 0 0
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exa_8_2.sce
//Exa 8.2 clc; clear; close; //given data t_hi=160;// in degree C t_ci=25;// in degree C t_ho=60;// in degree C Mh=2;// in kg/s Mc=2;// in kg/s Cph=2.035; // in kJ/kg degree C Cpc=4.187;// in kJ/kg degree C U=250;// in W/m^2 K d=0.5;// in m // Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci) t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci;// in degree C del_t1=t_hi-t_co;//in degree C del_t2=t_ho-t_ci;//in degree C del_tm= (del_t1-del_t2)/log(del_t1/del_t2); Cph=Cph*10^3;// in J/kg degree C q=Mh*Cph*(t_hi-t_ho); //Formula q=U*%pi*d*l*del_tm l=q/(U*%pi*d*del_tm); disp(l,"Length of the heat exchanger in meter")
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Example45.sce
// Display mode mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Turbomachinery Design and Theory,Rama S. R. Gorla and Aijaz A. Khan, Chapter 4, Example 5") disp("Impeller tip speed is given by") D = 0.914; N = 9000; U2 = %pi*D*N/60 disp("Since the exit is radial and no slip, Cw2 = U2 = 431 m/s") disp("From the velocity triangle,") alpha2 = 20; Cw2 = U2; Cr2 = U2*tan(alpha2 *%pi/180) disp("For radial exit, relative velocity is exactly perpendicular to rotational velocity U2. Thus the angle beta2 is 90degrees for radial exit.") disp("Using the velocity triangle") C2 = (U2^2 + Cr2^2)^0.5
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8.sce
// Caption: Finding per unit system clear; close; clc; Z_baseH=2400/20.8; Z_baseX=240/208; I_x=5.41/208;//per unit at low voltage side Z_eqH=(1.42+%i*1.82)/115.2;//per unit disp(Z_eqH,'equivalent impedence referred to high voltage side')
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W_5.sce
// sum 12-5 clc; clear; h=10; t=10/sqrt(2); Ta=80; x=((50*25)+(50*0))/(50+50); y=x; ra=sqrt(x^2+37.5^2); Ixx=(7.07*50^3/12)+(50*7.07*(12.5^2))+(50*7.07*12.5^2); IG=2*Ixx; e=100+(50-12.5); Tr=16.09*10^-3; P=Ta/Tr; P=P*10^-3; // printing data in scilab o/p window printf("P is %0.3f KN ",P);
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Ex3_43.sce
clc; t=1/2; //ratio of continuous rating to one hour rating p=2; // ratio of new KVA rating to old KVA rating al=2*(p*t); printf('Ratio of core loss to ohmic loss is %f ',al);
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Eigen.sce
clc;close;clear; function [A]=Eigen(A) lam = poly(0,'lam') lam = lam charMat = A-lam*eye(3,3) disp(charMat,'The characteristic matrix is') charPoly = poly(A,'lam') disp(charPoly,'the characteristic polynomial is') lam = spec(A) disp(lam,'the eigen values of A are') function[x,lam]=eigenvectors(A) [n,m]=size(A); lam=spec(A)'; x=[]; for k = 1:3 B=A-lam(k)*eye(3,3); C=B(1:n-1,1:n-1); b=B(1:n-1,n) y=C\b; y=[y;1]; y=y/norm(y); x=[x y]; end endfunction get('eigenvectors') [x,lam]=eigenvectors(A) disp(x,'The eigen vectors of A are'); endfunction function main() A=[0,0,0;0,0,0;0,0,0] A(1,1)=input("enter a11: ") A(1,2)=input("enter a12: ") A(1,3)=input("enter a13: ") A(2,1)=input("enter a21: ") A(2,2)=input("enter a22: ") A(2,3)=input("enter a23: ") A(3,1)=input("enter a31: ") A(3,2)=input("enter a32: ") A(3,3)=input("enter a33: ") [A]=Eigen(A); endfunction main();
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EX_2_38_a.sce
// Example 2.38.a:S clc; clear; close; Beta=180;//Common emitter D.C. Current gain Re=1;// Collector resistance in killo ohms R1=5.76;// resistance in killo ohms R2=34.67;// resistance in killo ohms S=1+Beta; disp(S,"Stability factor in fixed bias case is")
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Ex5_1.sce
//Continuous Time Fourier Series Coefficients of //a periodic signal x(t) = sin(2*Wot) clear; close; clc; t = 0:0.01:1; T = 1; Wo = 2*%pi/T; xt = sin(2*Wo*t); for k =0:4 C(k+1,:) = exp(-sqrt(-1)*Wo*t.*k); a(k+1) = xt*C(k+1,:)'/length(t); if(abs(a(k+1))<=0.01) a(k+1)=0; end end a =a'; ak = [-a,a(2:$)] for i=1:length(ak) if real(ak(i))== 0 then phase(i)=0; else if i<length(ak)/2 then phase(i)= atan(imag(ak(i))/real(ak(i))); else phase(i)= -atan(imag(ak(i))/real(ak(i))); end end end disp("The fourier series coefficients are...") disp(ak) disp("magnitude of Fourier series coefficient") disp(abs(ak)) disp("Phase of Fourier series coefficient in radians") disp(phase) n=-4:4; subplot(2,1,1) plot(n,abs(ak),'.'); xtitle("|ak|","k","|ak|"); subplot(2,1,2) for i=1:length(n) if n(i)== -2 then phase(i)=3.14/2; elseif n(i)== 2 then phase(i)= -3.14/2; else phase(i)=0; end end plot(n,phase,'.'); xtitle("/_ak","k","/_ak");
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EX2_3_a.sce
//ex_2.3.a even and odd signals of x(t) clear; clc; close; t = 0:0.01:5; x=exp(t) figure a=gca(); xtitle('x(t)') plot2d(t,x) figure a=gca(); xtitle('even signal') plot2d(t,x/2) t1=-5:1/100:0; plot2d(t1,x($:-1:1)/2) a.y_location='origin' figure a=gca(); xtitle('odd signal') plot2d(t,x/2) t1=-5:1/100:0; plot2d(t1,-x($:-1:1)/2) a.y_location='origin' a.x_location='origin'
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example_2_2.sce
syms t s w; y=laplace('sin(w*t)',t,s); disp(y,"ans=")
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ench12.tst
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//Chapter 10 //Example 10.2 //Page 256 //loadedfault clear;clc; //Given Pgm = 30e6; Vgm = 13.2e3; Xgm = 0.20; Xl = 0.10; Pm = 20e6;pfm = 0.8;Vt_m = 12.8e3; Pbase = Pgm; Vbase = Vgm; Vf = Vt_m / Vbase; Ibase = Pbase / (sqrt(3) * Vbase); I_L = (Pm / (pfm * sqrt(3) * Vt_m)) * (cos(36.9 * %pi/180) + %i * sin(36.9 * %pi / 180)) / Ibase; disp(I_L,'Line Current in per unit is') Vt_g = Vf + (%i * Xl) * I_L; E11_g = Vt_g + (%i * Xgm) * I_L; I11_g = E11_g / (%i * (Xgm + Xl)); I11_gA = Ibase * I11_g; disp(I11_g,'Fault current in the generator side in per unit') disp(I11_gA,'Fault current in the generator side in A') E11_m = Vf - (%i * Xgm) * I_L; I11_m = E11_m / (%i * (Xgm)); I11_mA = Ibase * I11_m; disp(I11_m,'Fault current in the motor side in per unit') disp(I11_mA,'Fault current in the motor side in A') If = I11_g + I11_m; disp(If,'Toatl Fault current in per unit') disp(If * Ibase,'Total Fault current in A')
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//Example 1_3_u2 clc(); clear; //To calculate the uncertenity in momentum h=6.63e-34 deltax=2*%pi*10^-9 deltap=h/(2*deltax) //units in Kg ms^-1 printf("The uncertenity in momentum is delta p=") disp(deltap) printf("Kg ms^-1") //In text book the answer is printed wrong as 0.53*10^-15 Kg ms^-1 the correct answer is 5.276D-26 Kg ms^-1
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function [out]=stack(n,varargin) //build model (cell)array by stacking models along array dimensions // //Calling Seqence //out=stack(n,sys1,sys2.....) // //Parameters //sys : siso or mimo lti system or system array(cell). //n : arraydimension //out:(cell)array of input systems //Description //out = stack(n,sys1,sys2,...) produces an array of dynamic system models // out by stacking (concatenating) the models (or arrays) sys1,sys2,... //along the array dimension. All system models must have the same number of inputs // and outputs (the same I/O dimensions) // //Examples //s=%s; //sys1=s/(s*s+5s+1) //sys2=s/(s+1) //out1=stack(1,sys1,sys2) //out1{:,:} //out2=stack(3,sys1,sys2) //out2{:,:} // //Author //Ayush Kumar /////////////////////////////////////////////////////////////////////////////////////// [lhs,rhs]=argn(0) for i=1:rhs-1 if (typeof(varargin(i))~="state-space"&&typeof(varargin(i))~="rational"&&typeof(varargin(i))~="polynomial"&&typeof(varargin(i))~="ce") then error(msprintf(gettext("%s :input argument should be a lti sysytem or lti system array"),"stack")) end, end, flag3=0 for i=1:rhs-1 if typeof(varargin(i))=="ce" then flag3=1 end, end if flag3==1 then for i=1:rhs-1 //in case of a cell array input ,all inputs should be a cell array if typeof(varargin(i))~="ce" then error(msprintf(gettext("%s:wrong type of input argument input"),"stack")) end end end, count=0; for i=1:rhs-1 if typeof(varargin(i))=="rational" || typeof(varargin(i))=="polynomial" then count=count+1; end, if typeof(varargin(i))=="state-space" then count=count+2; end, if typeof(varargin(i))=="ce" then count=count+3; end, end, /////pure rational or polynomial input models if count==rhs-1 then for i=1:rhs-1 testmat{i,1}=varargin(i); //storing data in a cell end end /////////mixture state-space and rational input models/////////////// if count>(rhs-1) then for i=1:rhs-1 //if flag==1 then if typeof(varargin(i))=="rational" || typeof(varargin(i))=="polynomial" then testmat{i,1}=tf2ss(varargin(i)) //converting to state-space elseif typeof(varargin(i))=="state-space" then testmat{i,1}=varargin(i); elseif typeof(varargin(i))=="ce" then tempcell=varargin(i); for j=1:prod(size(tempcell)) if typeof(tempcell{j})=="rational" || typeof(tempcell{j})=="polynomial" then k=tf2ss(tempcell{j}) tempcell{j}=k end end testmat{i,1}=tempcell; end end //k=testmat{1,1} //[nxip,nuip]=size(k{1,1}) //[nxop,nuop]=size(testmat{1,1}.C) for i=1:rhs-1 if typeof(testmat{i,1})=="state-space" then [nxip,nuip]=size(testmat{1,1}) if or(size(testmat{i,1})<>[nxip,nuip]) then error(msprintf(gettext("%s:input systems should have same input output matrix sizes"),"stack")) end end if typeof(testmat{i,1})=="ce" then //k=testmat{1,1} //[nxip,nuip]=size(k{1,1}) for j=1:prod(size(testmat{i,1}))-1 tempcell=testmat{i,1}; if or(size(tempcell{j})<>size(tempcell{j+1})) then error(msprintf(gettext("%s:input systems should have same input output matrix sizes"),"stack")) end, end, end, end, end, mat=[] for i=1:n-1 mat(i)=1 end, mat(n)=rhs-1; if n==1 then tempmat=cell(mat,n); else tempmat=cell(mat); end if n==1 then for i=1:rhs-1 for j=1:n //adding system in tempmat along array dimension tempmat{i,j}=testmat{i+j-1,1} end, end, else for i=1:rhs-1 tempmat{$,i}=testmat{i,1}; end, end out=tempmat; endfunction //out{:,:}
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//Example 12.17 //Program to calculate: //(a)The optimium receiver bandwidth //(b)The peak to peak signal power to rms noise ratio clear; clc ; close ; //Given data Tr=12*10^(-9); //s - SYSTEM RISE TIME fo=20*10^6; //Hz - NOMINAL PULSE RATE fd=5*10^6; //Hz - PEAK TO PEAK FREQUECY DEVIATION M=60; //APD MULTIPLICATION FACTOR R=0.7; //APD RESPONSIVITY B=6*10^6; //Hz - BASEBAND NOISE BANDWIDTH Ppo=10^(-7); //Watt - PEAK OPTICAL POWER in_sq_bar=10^(-17); //A^2 - RECEIVER MEAN SQUARE NOISE CURRENT //(a)The optimium receiver bandwidth Bopt=1/Tr; To=1/fo; //(b)The peak to peak signal power to rms noise ratio SNR=3*(To*fd*M*R*Ppo)^2/((2*%pi*Tr*B)^2*in_sq_bar); //Displaying the Results in Command Window printf("\n\n\t (a)The optimium receiver bandwidth is %0.1f MHz.",Bopt/10^6); printf("\n\n\t (b)The peak to peak signal power to rms noise ratio is %0.1f dB.",10*log10(SNR));
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// Demo script for linear regression getd('../') data_url = 'https://raw.githubusercontent.com/franklinwillemen/Machine_Learning/master/Regression/Simple_Linear_Regression/Salary_Data.csv' machineLearnURLDownload(data_url) machineLearnCustomURL('custom', 'preprocessing.py');
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// EJERCICIO 1 Determine gráficamente valores aproximados de las primeras tres raíces positivas de f(x)=cos(x)cosh(x)+1 deff ('y = f(x)', 'y = cos(x).* cosh(x)+1'); x=0:.01:15; // el 15 sale de ir probando a ver donde estan las raíces. a=gca(); // declarar un manejador de ejes a.x_location="middle"; // acceder a una propiedad de un objeto (en este caso a un campo de a, x location) // otra forma de hacer que represente el x=0 sin gca es plot(x,f(x),x,0) plot (x,f(x)); disp("las raíces son 8, 11 , 14.14"); //MÉTODO DE LA BISECCIÓN. Converge siempre a una raíz (convergencia global). Para elegir el intervalo donde quiero aplicarlo podemos usar este tipo de gráficas._
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//Problem 20.24: An a.c. source of 24 V and internal resistance 15 kohm is matched to a load by a 25:1 ideal transformer. Determine (a) the value of the load resistance and (b) the power dissipated in the load. //initializing the variables: tr = 25; // teurn ratio V = 24; // in Volts R1 = 15000; // in Ohms Rin = 15000; // in ohms //calculation: //Turns ratio, tr = N1/N2 = V1/V2 //For maximum power transfer R1 needs to be equal to 15 kohm RL = R1/(tr^2) //The total input resistance when the source is connected to the matching transformer is Rt = Rin + R1 //Primary current, I1 = V/Rt //N1/N2 = I2/I1 I2 = I1*tr //Power dissipated in load resistor RL P = I2*I2*RL printf("\n\n Result \n\n") printf("\n (a) the load resistance is %.0f ohm", RL) printf("\n (b) power dissipated in the load resistor is %.2E W", P)
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// Example 5.5 page no-287 clear clc b=50 //Beta vcc=10 //V rc= 250 //ohm ib=0.4 //mA ic=21 //mA vce=vcc-((ic+ib)*rc/1000) vce=floor(vce*10)/10//aproximated to printf("\nVce = %.1fV",vce) vbe=0.6 rb=(vce-vbe)/ib s=(b+1)/(1+(b*rc/(rc+rb*1000))) printf("\nRb = %.0f K-Ohm\nS = %d",rb,ceil(s))
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clear; clc; // Implementação resolução pelo método de Jacobi function [x, Er]= jacobi(A,b,x0,n) [l,c]=size(A) erro = 1; cont = 1; x = x0; while (cont <= n) xa = x for i = 1:l soma=0; for j = 1:l if(j~=i) then soma = soma + A(i,j)*xa(j); end end x(i) = (b(i) - soma)/A(i,i); end Er(cont) = max(abs(x-xa))/max(abs(x)) cont = cont + 1; end endfunction //Declarando as variáveis que serão utilizadas A = [15 5 -5;1 10 1;2 -2 8]; b = [30 23 -10]'; x0 = zeros(1,3); n = 10; //Exemplo chamada [x, Er]= jacobi(A,b,x0,n);
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//Solution 5-06 WD=get_absolute_file_path('5_06_solution.sce'); datafile=WD+filesep()+'5_06_example.sci'; clc; exec(datafile) V_2 = sqrt(2 * g * z_1); //Toricelli equation printf("Water leaves the tank with initial velocity of %1.2f m/s", V_2);
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clc //initialisation of variables d= 0.275 //in p= 15 p1= 20 p3= 8 //CALCULATIONS Fs= (d*p/100)+(d*p1/100)-(d*p3/100) Fs1= Fs*100/d //RESULTS printf ('final available squeeze = %.2f percent',Fs1)
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//scilab 5.4.1 //Windows 7 operating system //chapter 14 Integrated Circuits and Operational Amplifiers clc clear R1=1*1000//R1=input resistance in ohms in the inverting amplifier circuit Rf=50*1000//Rf=feedback resistance in ohms A=-(Rf/R1)//AV=voltage gain of the inverting amplifier circuit disp(A,"The voltage gain of the given inverting amplifier circuit is =") //vin=0.5*sin(100*%pi*t) //vout=A*vin=-50*0.5*sin(100*%pi*t)=-25*sin(100*%pi*t) disp("If the operation were entirely linear ,the output voltage would have been -25*sin(100*%pi*t)") disp("But since the voltage supply is +-12V ,the op-amp is saturated when |vout| attains 12V") //Let at time t=to,vout=-12V //-12=-25*sin(100*%pi*to) to=(1/(100*%pi))*asin(12/25) format("v",8) disp("s",to,"to=") disp("Thus over the entire cycle,") disp("vout=-25*sin(100*%pi*t) V when 0<=t<=to") disp("vout=-12V when to<=t<=(0.01-to)") disp("vout=-25*sin(100*%pi*t) V when (0.01-to)<=t<=(0.01+to)") disp("vout=+12V when (0.01+to)<=t<=(0.02-to)") disp("vout=-25*sin(100*%pi*t) V when (0.02-to)<=t<=0.02 seconds")
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// Example 13.6 // Inversion with complex Poles s=%s; t=0:0.001:10 num=15*s^2-16*s-7; den=(s+2)*(s^2+6*s+25); F_s=num/den; pfe=pfss(F_s); // partial fraction of the transfer function // from pfe(1) we get B=10; C=-66; alpha=3;// from pfe(1) beta=sqrt(25-9);//Comparing the denominator of pfe(1) with standard 2nd orderequation // Now K=B+(%i*(alpha*B-C))/beta; // From inverse Laplace Transfrom of pfe(2) we get f1=5*exp(-2*t) K_m=abs(K); // Magnitude of K phase_K=atan(imag(K),real(K)); g=K_m*exp(-alpha*t).*cos(beta*t+phase_K); f=f1+g; plot(t,f) xlabel('t') ylabel('f(t)') title('Function Waveform')
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clc //Example 16.4 disp('Given') disp('R1=2 ohm R2=3 ohm L=1H C=125mF') R1=2;R2=3 ; L=1;C=125*10^-3; w0=sqrt(1/(L*C)-(R1/L)^2) printf("w0=%d rad/s \n",w0) //Input admittance is 1/R2+i*w*C+1/(R+I*w*L) Y=1/3+%i/4+1/(2+%i*2) printf("Y= %3.4f S \n",Y) //Now input impedance at resonance Z=1/Y printf("Z= %3.4f ohm \n",Z) //Resonant frequency f=1/sqrt(L*C) f=1/sqrt(L*C) printf("f=%3.2f rad/s \n",f);
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//exapple 1.31 clc; funcprot(0); // Initialization of Variable longP=112+20/60+15/3600;//longitude of place GST=8+10/60+28/3600;//GST at GMM lst=18+28/60+12/3600;//local sidereal time dot=longP/15;//difference in time gmm=lst+dot-GST;//SI at GMM i=gmm*9.8565/3600;//error gmm=gmm-i;//LST at L.M.N LMT=gmm-dot;//local mean time disp("local mean time in past midnight observed:"); a=modulo(LMT*3600,60); printf("seconds %.2f",a); b=modulo(LMT*3600-a,3600)/60; printf(" minutes %i",b); c=(LMT*3600-b*60-a)/3600; if c>24 then c=c-24; end printf(" hours %i",c);
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clc // Given that V = 30e3 // voltage in V lambda_min = 0.414e-10 // shortest wavelength in m e = 1.6e-19 // charge on an electron in C c = 3e8 // speed of light in m/sec // Sample Problem 2 on page no. 20.7 printf("\n # PROBLEM 2 # \n") printf("Standard formula used \n ") printf("h*c/lambda = eV \n") h = (e * V * lambda_min) / c printf("\n Planck constant is %e J sec.",h)
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loadmatfile("matriz.txt") //Funcao que calcula a linha com maior numero de zeros nlin = 0 cont2 = 0 for i = 1:4 cont = 0 for j = 1:4 if matriz(i,j) == 0 then cont = cont + 1 end end if cont > cont2 then nlin = i cont2 = cont for j = 1:4 linha(1,j) = matriz(i,j) end end end //Caso não houver zero em nenhuma linha é escolhida a primeira linha if nlin == 0 then nlin = 1 for j = 1:4 linha(1,j) = matriz(1,j) end end //Funções que calculam as matrizes que se formam pelo uso do método do Laplace //Caso a linha com maior numero de zeros seja a primeira if nlin == 1 then for i = 1:3 for j = 1:3 matriza(i,j) = matriz(i+1,j+1) if j == 1 then matrizb(i,j) = matriz(i+1,j) else matrizb(i,j) = matriz(i+1,j+1) end if j == 3 then matrizc(i,j) = matriz(i+1,j+1) else matrizc(i,j) = matriz(i+1,j) end matrizd(i,j) = matriz(i+1,j) end end end //Caso a linha com maior numero de zeros seja a segunda if nlin == 2 then for i = 1:3 for j = 1:3 if i == 1 then matriza(i,j) = matriz(i,j+1) if j == 1 then matrizb(i,j) = matriz(i,j) else matrizb(i,j) = matriz(i,j+1) end if j == 3 then matrizc(i,j) = matriz(i,j+1) else matrizc(i,j) = matriz(i,j) end matrizd(i,j) = matriz(i,j) else matriza(i,j) = matriz(i+1,j+1) if j == 1 then matrizb(i,j) = matriz(i+1,j) else matrizb(i,j) = matriz(i+1,j+1) end if j == 3 then matrizc(i,j) = matriz(i+1,j+1) else matrizc(i,j) = matriz(i+1,j) end matrizd(i,j) = matriz(i+1,j) end end end end //Caso a linha com maior numero de zeros seja a terceira if nlin == 3 then for i = 1:3 for j = 1:3 if i == 3 then matriza(i,j) = matriz(i+1,j+1) if j == 1 then matrizb(i,j) = matriz(i+1,j) else matrizb(i,j) = matriz(i+1,j+1) end if j == 3 then matrizc(i,j) = matriz(i+1,j+1) else matrizc(i,j) = matriz(i+1,j) end matrizd(i,j) = matriz(i+1,j) else matriza(i,j) = matriz(i,j+1) if j == 1 then matrizb(i,j) = matriz(i,j) else matrizb(i,j) = matriz(i,j+1) end if j == 3 then matrizc(i,j) = matriz(i,j+1) else matrizc(i,j) = matriz(i,j) end matrizd(i,j) = matriz(i,j) end end end end //Caso a linha com maior numeros de zeros seja a quarta if nlin == 4 then for i = 1:3 for j = 1:3 matriza(i,j) = matriz(i,j+1) if j == 1 then matrizb(i,j) = matriz(i,j) else matrizb(i,j) = matriz(i,j+1) end if j == 3 then matrizc(i,j) = matriz(i,j+1) else matrizc(i,j) = matriz(i,j) end matrizd(i,j) = matriz(i,j) end end end soma = 0 //Atribuindo determinantes a um vetor determinantes(1,1) = det(matriza) determinantes(1,2) = det(matrizb) determinantes(1,3) = det(matrizc) determinantes(1,4) = det(matrizd) //Calculo do Determinante pelo metodo Laplace for j = 1:4 if modulo(nlin+j,2) <> 0 then determinantes(1,j) = determinantes(1,j) * -1 end soma = soma + determinantes(1,j)*matriz(nlin,j) end //Calculo do determinantes por comando Scilab scilab = det(matriz) //Apresentação de dados disp(linha,"Linha escolhida: ") disp(nlin,"Numero da linha: ") disp(soma,"Determinantes da matriz pelo metodo do Laplace: ") disp(scilab,"Determinante calculado por comando scilab: ")
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// Scilab code Ex7.3: Pg.292 (2008) clc; clear; l_1 = 2; // Orbital quantum number l_2 = 3; // Orbital quantum number printf("\nThe possible values of l are:"); for l = (l_2-l_1):1:(l_1 + l_2) printf(" %d ", l); end; // Result // The possible values of l are: 1 2 3 4 5
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//Section-14,Example-1,Page no.-PC.7 //To find the temperature at which pressure of gas will reach the bursting value. clc; //PV=nRT P=10 V=(10^-3)*(1/10^-3) n=((5*10^-3)/30) R=0.0821 T=((P*V)/(n*R)) disp(T,'Required temperature(K)')
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clc; clear all; disp("heat lost by man") d=0.35;// m h=1.65;//m ts=28;// degree C ta=12;// degree C U=30*1000/3600;// m/s tf=(ts+ta)/2; // film temperature k=2.59*10^(-2);// W/m.C v=15*10^(-6);// m^2/s Pr=0.707; Re=U*d/v; disp("Nu=C*Re^n*Pr^(1/3)") C=0.027; n=0.805; Nu=C*Re^n*Pr^(1/3); hs=Nu*k/d Q=hs*%pi*d*h*(ts-ta); disp("w",Q,"heat lost by man =")
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clc; x=0.9; vg=0.1104; v=x*vg; disp("specific volume is:"); disp("m^3/kg",v) hf=885; h_fg=1912; h=hf+x*h_fg; disp("specific enthalpy is:"); disp("kJ/kg",h); uf=883; ug=2598; u=(1-x)*uf+x*ug; disp("specific internal energy is:"); disp("kJ/kg",u);
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// Example 7.14 Apply the geometric mean to find general index from the following clc; clear; I=[118 120 97 107 111 93]; W=[4 1 2 6 5 2]; IN=sum(log(I).*W)/sum(W); GI=exp(IN) disp(GI,"General Index",IN,"Log General Index number =")
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//Fx(xn+0.8h) = [C1Fn-1 + C2Fn + C3Fn+1] / H // ESSE AQUI A GENTE USA PRA DERIVADA PRIMEIRA HEEEIN x =[0 1/9 2] // corresponde a n+0,n + 1/9 e n + 2 xc= 0 // onde é calculada a derivada. (xn+0), ficaria = 0 b(1)=0 b(2)=1 b(3)=2*xc //tem que usar aqui xc for i=1:3 M(1,i)=1 M(2,i)=x(i) M(3,i)=x(i)^2 end c=inv(M)*b disp('Coeficientes: ') disp(c)
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//Chapter 14 //Example 14.3 //page 538 //To estimate the values of the random variables x1 and x2 using WLSE clear;clc; i=0; x=1;y=8.5 printf('---------------------------------------\n'); printf('iteration\t\tx(l)\n'); printf('---------------------------------------\n'); printf('\t%d\t\t%0.3f\n',i,x); for i=1:1:10 k=(1/3)*x^-2 //expression for the value of k has been printed wrongly in the textbook x=x+(k)*(y-x^3); printf('\t%d\t\t%0.3f\n',i,x); end
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Ex2_6.sce
clear all; clc; //This numerical is Ex 2_1S,page 29. Q1=18.2 N1=1000 N2=1500 delta_p1=10.3 P_s1=6 Q2=Q1*N2/N1 printf("\n The value of Q2 is equal to %g m^3/h",Q2) delta_p2=delta_p1*((N2/N1)^2) printf("\n The value of delta_p2 is equal to %0.1f bars",delta_p2) P_s2=P_s1*(N2/N1)^3 printf("\n The value of P_s2 is equal to %g kW",P_s2) E1=((Q1/3600)*delta_p1*10^2)/(P_s1) printf("\n The value of E1=E2 is equal to %g ",E1) disp("Thus the efficiency is equal to 86.8%")
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/1691/CH2/EX2.34/exmp2_34.sce
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exmp2_34.sce
//Example 2.34 clc disp("The name of the oscillator is Pierce oscillator") disp("C1 = 1000 pF, C2 = 100 pF, f_s = 1 MHz") ceq=(1000*100*10^-12)/1100 format(11) disp(ceq,"C_eq(in F) = C1*C2 / C1+C2 =") disp("At resonance, X_L = X_Ceq i.e. 2*pi*f*L = 1 / 2*pi*f*C_eq") l=(1/(((2*%pi*10^6)^2)*(90.909*10^-12)))*10^6 format(4) disp(l,"Therefore, L(in uH) = 1/(2*pi*f)^2*C_eq =") disp("The fig 2.61(a) shows the electrical equivalent of the crystal") disp("At series resonance,") disp("X_L = X_C for crystal") disp("Therefore, C = 90.909 pF for crystal") disp("The mounting capacitance is about 1 to 2 pF")
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/1592/CH2/EX2.10/example_2_10.sce
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example_2_10.sce
//Scilab Code for Example 2.10 of Signals and systems by //P.Ramakrishna Rao //The value of X(s) is found by solving the differential equation clear; clc; syms t s; s= %s; X=pfss((s^2+8*s+6)/((s+2)*(s+3)*s)); X(1)=1/s; f1=ilaplace(X(1)) f2=ilaplace(X(2)) f3=ilaplace(X(3)) fz=f1+f2+f3; disp(fz*'u(t)',"c) x(t)=");
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/2201/CH3/EX3.22/ex3_22.sce
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ex3_22.sce
// Exa 3.22 clc; clear; close; // Given data h = 6.64*10^-34;// in J-s q=1.6*10^-19;// in C h= h/q;// in eV c = 3*10^8;// in m/s lembda = 0.87*10^-6;// in m E_g = (h*c)/lembda;// in eV disp(E_g,"The band gap in eV is");